Calculus rests on one idea: what a function approaches, whether or not it ever gets there. Grade 12 introduced limits by substitution; here you handle the harder cases and meet the precise definition.
🎯 What a limit says
limx→a f(x) = L means f(x) can be made as close to L as we like by taking x close enough to a, from both sides. The value f(a) plays no part: it can differ from L or not exist at all.
The one-sided limits limx→a⁻ and limx→a⁺ approach from the left and the right. The two-sided limit exists only when both exist and are equal.
f(x) = (x² − 4)/(x − 2) is undefined at x = 2
For x ≠ 2: f(x) = (x − 2)(x + 2)/(x − 2) = x + 2
lim as x→2 = 2 + 2 = 4
🖼️ See it
y = (x² − 4)/(x − 2) is the line y = x + 2 with a hole at x = 2. The limit there is still 4.
🖼️ See it
The left limit at x = 2 is 1 and the right limit is 3, so the two-sided limit does not exist.
🧰 Techniques for 0/0
If direct substitution gives a number, that is the limit (for polynomials, rational functions, roots and trig functions where they are defined). If it gives 0/0, the limit is undecided and the expression must be rewritten:
Factor and cancel.
Rationalise: multiply top and bottom by the conjugate.
Combine fractions over a common denominator.
lim as x→0 of (√(x + 4) − 2)/x
Multiply by (√(x + 4) + 2)/(√(x + 4) + 2):
= x / [x(√(x + 4) + 2)] = 1/(√(x + 4) + 2)
→ 1/(2 + 2) = 1/4
🥪 The Squeeze Theorem and sin x / x
If g(x) ≤ f(x) ≤ h(x) near a, and g and h both approach L, then f must approach L too. That argument proves the most important trig limit (x in radians):
lim as x→0 of sin(x)/x = 1
lim as x→0 of sin(5x)/x = 5 · lim sin(5x)/(5x) = 5
Squeeze: −x² ≤ x² sin(1/x) ≤ x², both sides → 0
so lim as x→0 of x² sin(1/x) = 0
🔬 The precise definition
The ε–δ definition turns “close” into numbers: for every ε > 0 there is a δ > 0 such that if 0 < |x − a| < δ, then |f(x) − L| < ε. Whatever accuracy ε is demanded of the output, some closeness δ of the input guarantees it.
Prove lim as x→3 of 2x = 6
|2x − 6| = 2|x − 3|, so 2|x − 3| < ε when |x − 3| < ε/2
Choose δ = ε/2
Common mistake. Writing “0/0” as an answer, or concluding that the limit does not exist. 0/0 only means this method has not finished yet.
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Meaning
lim f(x) as x→2 describes:
Does f need to be defined at 2?
✍️ Type your answer · 2 of 8
Find the limit as x→3 of (x² − 9)/(x − 3).
What is the limit?
Enter a whole number or a fraction such as -3/4.
Factor the top as a difference of squares.
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Find the limit as x→4 of (x² − x − 12)/(x − 4).
What is the limit?
Enter a whole number or a fraction such as -3/4.
x² − x − 12 has x − 4 as a factor.
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Find the limit as x→0 of (√(x + 9) − 3)/x.
What is the limit?
Enter a whole number or a fraction such as -3/4.
Multiply top and bottom by √(x + 9) + 3.
✍️ Type your answer · 5 of 8
Find the limit as x→0 of sin(5x)/x, with x in radians.
What is the limit?
Enter a whole number or a fraction such as -3/4.
Make the angle and the denominator match.
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Find the limit as x→2 of (1/x − 1/2)/(x − 2).
What is the limit?
Enter a whole number or a fraction such as -3/4.
Combine the two fractions on top first.
✏️ Quick check · 7 of 8
One-sided
The left limit at x = 1 is 3 and the right limit is 5. The limit at x = 1:
Compare the two sides.
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Indeterminate
Direct substitution gives 0/0. This means:
Think of (x² − 4)/(x − 2) at x = 2.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
The limit of sin(x)/x as x→0 (radians) is:
Question 2 of 5
To evaluate (√(x + 1) − 1)/x as x→0, the best first step is:
Question 3 of 5
The limit of (x² − 1)/(x − 1) as x→1 is:
Question 4 of 5
In the ε–δ definition, ε measures closeness of:
Question 5 of 5
If −x² ≤ f(x) ≤ x² near 0, then the limit of f at 0 is:
Lesson 2 · Limits
〰️ Continuity & Limits at Infinity
A continuous function has no breaks, and that simple property guarantees solutions exist. Limits at infinity then describe how a function behaves far from the origin.
〰️ Continuity at a point
f is continuous at a when three things hold: f(a) is defined, the limit as x→a exists, and the two are equal.
Removable discontinuity: a hole; the limit exists but f(a) is missing or different.
Jump: the one-sided limits exist but differ.
Infinite: the function grows without bound, as at a vertical asymptote.
f(x) = x² + k for x < 2, f(x) = 3x for x ≥ 2
Left limit: 4 + k. Right limit and f(2): 6
Continuous when 4 + k = 6, so k = 2
🖼️ See it
y = (2x + 1)/(x − 1) has a vertical asymptote at x = 1 and a horizontal asymptote at y = 2.
🖼️ See it
f(x) = x³ − x − 1 goes from −1 to 5 on [1, 2], so its continuous graph must cross zero.
🌉 The Intermediate Value Theorem
If f is continuous on [a, b] and N lies between f(a) and f(b), then f(c) = N for some c in (a, b). Its main use is showing that an equation has a root when a function changes sign.
f(x) = x³ − x − 1 is continuous on [1, 2]
f(1) = −1 and f(2) = 5
The signs differ, so x³ − x − 1 = 0 has a root in (1, 2)
What it does not say. The theorem proves a root exists but does not find it, and without continuity it fails: 1/x changes sign on [−1, 1] yet is never 0.
♾️ Limits at infinity and asymptotes
For a rational function as x→±∞, divide top and bottom by the highest power of x in the denominator. The degrees decide the result:
Top degree smaller: the limit is 0.
Degrees equal: the limit is the ratio of leading coefficients, giving a horizontal asymptote.
Top degree larger: no finite limit.
A vertical asymptote occurs where the denominator is 0 and the numerator is not, once common factors are cancelled.
(3x² + 1)/(x² − 5) = (3 + 1/x²)/(1 − 5/x²) → 3 as x→∞
√(x² + 4x) − x, as x→∞: multiply by the conjugate
= 4x/(√(x² + 4x) + x) → 4/(1 + 1) = 2
✍️ Type your answer · 1 of 8
Find the limit as x→∞ of (3x² + 1)/(x² − 5).
What is the limit?
Enter a whole number or a fraction such as -3/4.
Compare the highest powers.
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Infinite limits
As x→0, the function 1/x²:
Try x = 0.1 and x = −0.1.
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f(x) = x² + k for x < 2, and f(x) = 3x for x ≥ 2.
What value of k makes f continuous at x = 2?
Enter a whole number or a fraction such as -3/4.
Set the left limit equal to f(2).
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Asymptotes
The vertical asymptote of (x − 3)/(x² − 9) is at:
Cancel common factors first.
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y = (4x − 1)/(2x + 5).
The horizontal asymptote is y = what?
Enter a whole number or a fraction such as -3/4.
Equal degrees: divide the leading coefficients.
✏️ Quick check · 6 of 8
Intermediate Value Theorem
f is continuous, f(1) = −1 and f(2) = 5. It follows that:
Which value lies between −1 and 5?
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Discontinuities
At x = 3, the function (x² − 9)/(x − 3) has:
Does the limit exist?
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Find the limit as x→∞ of √(x² + 4x) − x.
What is the limit?
Enter a whole number or a fraction such as -3/4.
Multiply by the conjugate over itself.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
Continuity at a requires:
Question 2 of 5
The limit of (2x + 7)/(x² + 1) as x→∞ is:
Question 3 of 5
One-sided limits that exist but differ give a:
Question 4 of 5
Why does the IVT not apply to 1/x on [−1, 1]?
Question 5 of 5
The horizontal asymptote of (5x³ − x)/(2x³ + 4) is:
Lesson 3 · Derivatives
📐 The Derivative
The derivative is the instantaneous rate of change: the limit of the slopes of ever-shorter secant lines. It is defined by a limit, so everything from Lessons 1 and 2 applies.
📐 From secant to tangent
The average rate of change of f over [a, a + h] is the secant slope (f(a + h) − f(a))/h. Let h→0 and it becomes the derivative:
f′(a) = lim as h→0 of [f(a + h) − f(a)] / h
f(x) = x²
[(a + h)² − a²]/h = (2ah + h²)/h = 2a + h → 2a
So f′(x) = 2x
The tangent line at x = a is y = f(a) + f′(a)(x − a).
🖼️ See it
At x = 1 on y = x², the secant to x = 2 has slope 3. As the second point slides in, the slope approaches the tangent slope, 2.
🖼️ See it
y = |x| is continuous at 0 but its slope jumps from −1 to 1, so there is no single tangent.
🔤 Notation and meaning
f′(x), dy/dx and df/dx all name the same function. If s(t) is position, then s′(t) is velocity and s″(t) is acceleration. The units of a derivative are output units per input unit: metres per second, dollars per item.
Tangent to y = x² at x = 3
Point: (3, 9). Slope: f′(3) = 6
y = 9 + 6(x − 3), which is y = 6x − 9
🚧 Where derivatives fail
Differentiable implies continuous, but not the reverse. A continuous function has no derivative at:
a corner, such as |x| at 0, where the left slope is −1 and the right slope is 1;
a vertical tangent, such as the cube root of x at 0;
any discontinuity.
Exam technique. When a question says “from first principles” or “using the definition”, you must show the limit of the difference quotient. Quoting a rule earns no marks.
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f(x) = x².
What is the average rate of change of f from x = 1 to x = 3?
Enter a whole number or a fraction such as -3/4.
Rise over run between the two points.
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f(x) = x². From the definition, f′(x) = 2x.
What is f′(3)?
Enter a whole number or a fraction such as -3/4.
Substitute into 2x.
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f(x) = x² + 3x. Use the difference quotient, which simplifies to 2x + 3 + h.
What is f′(1)?
Enter a whole number or a fraction such as -3/4.
Let h→0, then substitute x = 1.
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Definition
f′(a) is defined as the limit as h→0 of:
Slope is change in y over change in x.
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y = x³, whose derivative is 3x².
What is the slope of the tangent at x = 2?
Enter a whole number or a fraction such as -3/4.
Substitute x = 2 into the derivative.
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The tangent to y = x² at x = 3 is y = mx + c.
What is the y-intercept c?
Enter a whole number or a fraction such as -3/4.
Use y = f(a) + f′(a)(x − a).
✏️ Quick check · 7 of 8
Differentiability
At x = 0, the function |x| is:
Compare the slopes on each side.
✏️ Quick check · 8 of 8
Meaning
If s(t) is position in metres and t is in seconds, s′(t) is:
Check the units of the rate.
📝 Lesson quiz · 5 questions
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Re-read the lesson, then give it another go.
Question 1 of 5
A secant line’s slope gives:
Question 2 of 5
If f is differentiable at a, then f is:
Question 3 of 5
The cube root of x has no derivative at 0 because of:
Question 4 of 5
From the definition, the derivative of 5x is:
Question 5 of 5
The units of dC/dx, with C in dollars and x in items, are:
Lesson 4 · Derivatives
🧮 Differentiation Rules
Nobody differentiates long expressions from the definition. A short list of rules, each proved from the limit, handles nearly every function built from powers, exponentials, logarithms and trig functions.
➕ Power, sum and constant rules
d/dx xⁿ = n xⁿ⁻¹ for any real n
d/dx [c f] = c f′
d/dx [f + g] = f′ + g′
y = 4x³ − 2x + 7 y′ = 12x² − 2
y = √x = x^(1/2) y′ = 1/(2√x)
y = 1/x² = x⁻² y′ = −2x⁻³
Rewrite roots and reciprocals as powers first. Higher derivatives simply repeat the process: f″ is the derivative of f′.
Common mistake. (fg)′ is not f′g′. Test it: x · x = x² has derivative 2x, but 1 · 1 = 1.
🌊 Exponentials, logs and trig
These assume x is in radians. Learn the table in the diagram until it is automatic.
y = x ln x y′ = 1 · ln x + x · (1/x) = ln x + 1
y = sin x cos x y′ = cos²x − sin²x
y = tan x = sin x / cos x
y′ = (cos²x + sin²x)/cos²x = sec²x
The special role of e: eˣ is the one exponential function that is its own derivative. For other bases, d/dx aˣ = aˣ ln a.
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f(x) = 4x³ − 2x + 7.
What is f′(1)?
Enter a whole number or a fraction such as -3/4.
Differentiate each term.
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f(x) = (x² + 1)(x³ − 2).
What is f′(1)?
Enter a whole number or a fraction such as -3/4.
Use the product rule, then substitute.
✍️ Type your answer · 3 of 8
f(x) = (2x + 1)/(x − 1).
What is f′(2)?
Enter a whole number or a fraction such as -3/4.
Use the quotient rule: low d-high minus high d-low, over low squared.
✏️ Quick check · 4 of 8
Trig
The derivative of sin x cos x is:
Treat it as a product of two functions.
✏️ Quick check · 5 of 8
Logs
The derivative of x ln x is:
x times ln x is a product.
✏️ Quick check · 6 of 8
Trig
The derivative of tan x is:
It is always positive where defined.
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f(x) = 3eˣ + 2 cos x.
What is f′(0)?
Enter a whole number or a fraction such as -3/4.
e⁰ = 1 and sin 0 = 0.
✍️ Type your answer · 8 of 8
f(x) = x⁴.
What is f″(1), the second derivative at 1?
Enter a whole number or a fraction such as -3/4.
Differentiate twice.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The derivative of 1/x is:
Question 2 of 5
The derivative of √x is:
Question 3 of 5
The derivative of cos x is:
Question 4 of 5
The derivative of 2ˣ is:
Question 5 of 5
The derivative of x² eˣ is:
Lesson 5 · Derivatives
⛓️ Chain Rule & Implicit Differentiation
Most real functions are compositions: a function inside a function. The chain rule handles those, and the same idea lets you differentiate equations you cannot solve for y.
⛓️ The chain rule
If y = f(g(x)), differentiate the outer function, leave the inside alone, then multiply by the derivative of the inside: dy/dx = f′(g(x)) · g′(x). Equivalently, dy/dx = (dy/du)(du/dx).
y = (x² + 1)⁵ y′ = 5(x² + 1)⁴ · 2x = 10x(x² + 1)⁴
y = sin(3x) y′ = 3 cos(3x)
y = e^(x²) y′ = 2x e^(x²)
y = √(x² + 9) y′ = x/√(x² + 9)
y = ln(cos x) y′ = −sin x / cos x = −tan x
Common mistake. Forgetting the inner derivative: the derivative of sin(3x) is not cos(3x).
🖼️ See it
🔁 Implicit differentiation
When x and y are tangled together, differentiate both sides with respect to x. Whenever you differentiate a term in y, the chain rule attaches a factor of dy/dx. Then solve for dy/dx.
d/dx arcsin x = 1/√(1 − x²)
d/dx arccos x = −1/√(1 − x²)
d/dx arctan x = 1/(1 + x²)
d/dx ln|x| = 1/x
For y = (function)^(function), take logs first. This is logarithmic differentiation:
y = xˣ, x > 0
ln y = x ln x
(1/y) y′ = ln x + 1
y′ = xˣ (ln x + 1)
✏️ Quick check · 1 of 8
Chain rule
The derivative of (x² + 1)⁵ is:
Outer derivative times inner derivative.
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f(x) = sin(3x).
What is f′(0)?
Enter a whole number or a fraction such as -3/4.
The inner function 3x has derivative 3.
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f(x) = e^(2x).
What is f′(0)?
Enter a whole number or a fraction such as -3/4.
Multiply by the derivative of 2x.
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f(x) = √(x² + 9).
What is f′(4)?
Enter a whole number or a fraction such as -3/4.
Inner derivative 2x cancels with the 2 from the root rule.
✏️ Quick check · 5 of 8
Chain rule
The derivative of ln(cos x) is:
d/dx ln u = u′/u.
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The circle x² + y² = 25.
What is dy/dx at the point (3, 4)?
Enter a whole number or a fraction such as -3/4.
Differentiate both sides, remembering the dy/dx on the y² term.
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The curve xy = 6.
What is dy/dx at the point (2, 3)?
Enter a whole number or a fraction such as -3/4.
Use the product rule on xy.
✏️ Quick check · 8 of 8
Inverse trig
The derivative of arctan x is:
It is defined for all real x.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The derivative of cos(x²) is:
Question 2 of 5
The derivative of y² with respect to x is:
Question 3 of 5
The derivative of arcsin x is:
Question 4 of 5
Logarithmic differentiation suits:
Question 5 of 5
The derivative of e^(sin x) is:
Lesson 6 · Applications
🪜 Related Rates & Linear Approximation
When quantities are linked by an equation, so are their rates of change. And near any point a differentiable function looks like its tangent line, which gives quick, accurate estimates.
🪜 Related rates
Differentiate an equation with respect to time, then substitute the values at the moment of interest. Substitute only after differentiating, because anything that is changing must stay a variable until then.
A 10 m ladder slides down a wall. The base moves out at 2 m/s.
How fast is the top moving when the base is 6 m out?
x² + y² = 100
2x (dx/dt) + 2y (dy/dt) = 0
At x = 6: y = 8
6(2) + 8(dy/dt) = 0
dy/dt = −1.5 m/s (negative: the top is falling)
🖼️ See it
🖼️ See it
Near x = 4, the tangent to y = √x is almost indistinguishable from the curve.
🌐 Shapes and rates
Circle: A = πr² dA/dt = 2πr (dr/dt)
Sphere: V = (4/3)πr³ dV/dt = 4πr² (dr/dt)
Cube: V = s³ dV/dt = 3s² (ds/dt)
Cars at a right angle: z² = x² + y²
z (dz/dt) = x (dx/dt) + y (dy/dt)
Signs matter. A quantity that is shrinking has a negative rate. Set this up before you substitute.
📏 Linear approximation
Near x = a, the tangent line is a good stand-in for f: L(x) = f(a) + f′(a)(x − a). In differential form, Δy ≈ f′(a) Δx.
Estimate √4.1 with f(x) = √x and a = 4
f(4) = 2 and f′(4) = 1/(2√4) = 1/4
L(4.1) = 2 + (1/4)(0.1) = 2.025
The true value is 2.02485…
The estimate worsens further from a. Concavity tells you the direction of the error: the tangent lies above a concave-down curve, so this estimate is slightly too high.
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A square’s side grows at 3 cm/s.
How fast is its area growing, in cm²/s, when the side is 10 cm?
Enter a whole number or a decimal such as 2.5.
Differentiate A = s² with respect to t.
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A cube’s edge grows at 2 cm/s.
How fast is its volume growing, in cm³/s, when the edge is 5 cm?
Enter a whole number or a decimal such as 2.5.
Differentiate V = s³ with respect to t.
✍️ Type your answer · 3 of 8
A 10 m ladder’s base moves away from a wall at 2 m/s.
What is dy/dt for the top, in m/s, when the base is 6 m from the wall?
Enter a whole number or a decimal such as 2.5.
Use x² + y² = 100. A falling top has a negative rate.
✍️ Type your answer · 4 of 8
Two cars leave a junction at right angles. At one moment they are 3 km and 4 km away, moving at 30 km/h and 40 km/h.
How fast is the distance between them growing, in km/h?
Enter a whole number or a decimal such as 2.5.
Use z² = x² + y².
✏️ Quick check · 5 of 8
Method
In a related rates problem, you substitute the known values:
What happens to the derivative of a number?
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Use the tangent to f(x) = √x at x = 4.
What is the linear estimate of √4.1?
Enter a whole number or a decimal such as 2.5.
L(x) = f(4) + f′(4)(x − 4).
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Use the tangent to f(x) = x³ at x = 2.
What is the linear estimate of 2.01³?
Enter a whole number or a decimal such as 2.5.
f(2) = 8 and f′(2) = 12.
✏️ Quick check · 8 of 8
Sign
Water drains from a tank at 5 litres per minute. With V the volume, dV/dt is:
Is V going up or down?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
For a circle with r = 5 and dr/dt = 2, dA/dt is:
Question 2 of 5
The linear approximation of f near a is:
Question 3 of 5
For a sphere, dV/dt equals:
Question 4 of 5
The tangent line estimate for a concave-down curve is:
Question 5 of 5
Differentials state that Δy is approximately:
Lesson 7 · Applications
📈 Curve Sketching
The first derivative shows where a function rises and falls. The second shows how it bends. Together they locate every important feature of a graph without plotting a single table of values.
🏔️ The first derivative
f′ > 0 means f is increasing; f′ < 0 means decreasing. A critical point is where f′ = 0 or f′ does not exist. Local extrema can occur only at critical points.
First derivative test: if f′ changes from + to − at c, it is a local maximum; from − to +, a local minimum; no sign change, neither.
f(x) = x³ − 3x
f′(x) = 3x² − 3 = 3(x − 1)(x + 1)
Critical points: x = −1 and x = 1
f′: + on x < −1, − on (−1, 1), + on x > 1
Local max at (−1, 2), local min at (1, −2)
🖼️ See it
f(x) = x³ − 3x: a maximum at x = −1, a minimum at x = 1, and an inflection point at x = 0.
🖼️ See it
🥣 The second derivative
f″ > 0 means concave up (a bowl); f″ < 0 means concave down (a dome). An inflection point is where concavity actually changes.
Second derivative test: with f′(c) = 0, f″(c) > 0 gives a local min and f″(c) < 0 a local max. If f″(c) = 0, the test tells you nothing: use the first derivative test.
f″(x) = 6x for f(x) = x³ − 3x
f″(−1) = −6: maximum. f″(1) = 6: minimum.
f″ changes sign at x = 0: inflection at (0, 0)
Common mistake. f″(c) = 0 does not guarantee an inflection point. For x⁴, f″(0) = 0, but x = 0 is a minimum.
🏁 Absolute extrema on a closed interval
A continuous function on [a, b] always has an absolute maximum and minimum (the Extreme Value Theorem). To find them, compare f at every critical point in the interval and at both endpoints.
f(x) = x³ − 3x on [0, 3]
Critical point in the interval: x = 1
f(0) = 0, f(1) = −2, f(3) = 18
Absolute min −2 at x = 1, absolute max 18 at x = 3
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f(x) = x³ − 12x.
At what positive x is there a critical point?
Enter a whole number or a fraction such as -3/4.
Set the derivative to zero.
✏️ Quick check · 2 of 8
First derivative
If f′(x) > 0 on an interval, then on that interval f is:
f′ is the slope.
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f(x) = x² − 6x + 1.
At what x is the minimum?
Enter a whole number or a fraction such as -3/4.
Solve f′(x) = 0.
✍️ Type your answer · 4 of 8
f(x) = x³ − 6x².
At what x is the inflection point?
Enter a whole number or a fraction such as -3/4.
Solve f″(x) = 0, then check the sign change.
✏️ Quick check · 5 of 8
Second derivative test
f′(c) = 0 and f″(c) < 0. At c, f has a:
Dome or bowl?
✍️ Type your answer · 6 of 8
f(x) = x³ − 3x on [0, 3].
What is the absolute maximum value?
Enter a whole number or a fraction such as -3/4.
Compare the critical point and both endpoints.
✍️ Type your answer · 7 of 8
f(x) = x³ − 3x on [0, 3].
What is the absolute minimum value?
Enter a whole number or a fraction such as -3/4.
Evaluate f at x = 1.
✏️ Quick check · 8 of 8
Inconclusive tests
For f(x) = x⁴, f′(0) = 0 and f″(0) = 0. At x = 0, f has a:
Look at the sign of f′ on each side.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
A critical point is where:
Question 2 of 5
An inflection point is where:
Question 3 of 5
f′ goes from − to + at c. Then c is a:
Question 4 of 5
Absolute extrema on [a, b] must be checked at critical points and:
Question 5 of 5
f″ > 0 on an interval means the graph is:
Lesson 8 · Applications
📦 Optimization
Largest area, least material, greatest profit: optimization turns a word problem into a function, then uses Lesson 7 to find its best value. The setup is the hard part.
📦 The method
Draw it and name the variables.
Write the quantity to optimize.
Use the constraint to reduce it to one variable.
State the domain the problem allows.
Differentiate and find critical points.
Justify that you have the maximum or minimum.
A box from a 12 cm square sheet, cutting squares of side x from the corners:
V = x(12 − 2x)², domain 0 < x < 6
V′ = (12 − 2x)(12 − 6x) = 0, so x = 2 (x = 6 is excluded)
V(2) = 2 · 8² = 128 cm³
V = 0 at both ends of the domain, so this is the maximum
🖼️ See it
V = x(12 − 2x)² on 0 < x < 6 rises to a peak of 128 at x = 2.
🖼️ See it
🌾 More classic setups
Fence three sides along a river, using 100 m:
2x + y = 100, so A = x(100 − 2x) = 100x − 2x²
A′ = 100 − 4x = 0, so x = 25, y = 50, A = 1250 m²
Revenue at price p = 60 − 2x for x items:
R = x(60 − 2x), R′ = 60 − 4x = 0, so x = 15
Closest point: minimise the squared distance.
It has the same minimum point, without the root.
Shortcut worth knowing. Squaring a positive quantity does not move its minimum, so d² gives the same answer as d with easier algebra.
✅ Proving it is the best
A critical point is only a candidate. Say which argument you use:
Closed interval: compare endpoints and critical points.
First derivative test: f′ changes from + to − gives the maximum if it is the only critical point in the domain.
Second derivative: f″ of one sign throughout the domain makes the critical point the absolute extreme.
Common mistakes. Forgetting the domain, answering with x when the question asks for the area or volume, and never showing that the point is a maximum.
✍️ Type your answer · 1 of 8
A rectangle has a perimeter of 40 m.
What is the largest possible area, in m²?
Enter a whole number or a decimal such as 2.5.
Use the perimeter to eliminate one side.
✍️ Type your answer · 2 of 8
A box is made by cutting squares of side x from the corners of a 12 cm sheet.
What value of x gives the greatest volume?
Enter a whole number or a decimal such as 2.5.
V = x(12 − 2x)², with 0 < x < 6.
✍️ Type your answer · 3 of 8
100 m of fence encloses three sides of a rectangle, with a river as the fourth side.
What is the largest area, in m²?
Enter a whole number or a decimal such as 2.5.
Two sides of x and one of 100 − 2x.
✍️ Type your answer · 4 of 8
Two numbers add to 10.
What is the smallest possible value of the sum of their squares?
Enter a whole number or a decimal such as 2.5.
Write everything in terms of one number.
✍️ Type your answer · 5 of 8
f(x) = x + 16/x for x > 0.
What is the minimum value of f?
Enter a whole number or a decimal such as 2.5.
Solve f′(x) = 0 first.
✏️ Quick check · 6 of 8
Justifying
After finding a critical point you should:
Could it be a minimum?
✍️ Type your answer · 7 of 8
At price 60 − 2x, a shop sells x items, so revenue is R = x(60 − 2x).
What value of x gives the greatest revenue?
Enter a whole number or a decimal such as 2.5.
Expand, then differentiate.
✏️ Quick check · 8 of 8
Setup
To find the point on a curve closest to the origin, it is easiest to minimise:
What makes the derivative simpler?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The constraint in an optimization problem is used to:
Question 2 of 5
Of all rectangles with a fixed perimeter, the greatest area belongs to:
Question 3 of 5
If f″ > 0 throughout the domain, the only critical point is:
Question 4 of 5
The domain of a box problem usually comes from:
Question 5 of 5
A question asks for the maximum area and you found x = 25. You should:
Lesson 9 · Theory
🧲 Mean Value Theorem & L’Hôpital’s Rule
The Mean Value Theorem is the bridge between a function and its derivative: many results in calculus are proved with it. L’Hôpital’s rule is a practical tool for limits that resist algebra.
🧲 The Mean Value Theorem
If f is continuous on [a, b] and differentiable on (a, b), then for some c in (a, b):
f′(c) = [f(b) − f(a)] / (b − a)
f(x) = x² on [1, 3]
Average slope: (9 − 1)/2 = 4
2c = 4, so c = 2
Somewhere the instantaneous rate equals the average rate. If you average 100 km/h on a trip, at some moment your speedometer read exactly 100.
Rolle’s theorem is the special case f(a) = f(b): then f′(c) = 0 for some c.
🖼️ See it
On [0, 2], y = x² has average slope 2. At c = 1 the tangent has that same slope, so it runs parallel to the secant.
🖼️ See it
🔗 Why it matters
The MVT proves facts that are easy to believe but need an argument:
If f′ = 0 on an interval, then f is constant.
If f′ = g′ on an interval, then f and g differ by a constant. This is the reason for the + C in Lesson 10.
If f′ > 0 on an interval, then f is increasing.
Check the hypotheses. |x| on [−1, 1] has average slope 0, yet f′ is never 0. The theorem does not apply because of the corner at 0.
🏥 L’Hôpital’s rule
If the limit of f/g has the form 0/0 or ∞/∞, then lim f/g = lim f′/g′, provided the second limit exists. Differentiate top and bottom separately; this is not the quotient rule.
lim as x→0 of (eˣ − 1)/x: 0/0, so → eˣ/1 → 1
lim as x→0 of (1 − cos x)/x²: 0/0, so → sin x/(2x)
still 0/0, so → cos x/2 → 1/2
lim as x→0⁺ of x ln x: rewrite as ln x/(1/x), which is ∞/∞
→ (1/x)/(−1/x²) = −x → 0
Forms such as 0 · ∞ and 1^∞ must first be rewritten as a fraction, often using logarithms.
✍️ Type your answer · 1 of 8
f(x) = x² on [1, 3].
What value of c satisfies the Mean Value Theorem?
Enter a whole number or a fraction such as -3/4.
Find the average slope, then set f′(c) equal to it.
✍️ Type your answer · 2 of 8
f(x) = x² + x on [0, 4].
What value of c satisfies the Mean Value Theorem?
Enter a whole number or a fraction such as -3/4.
f(0) = 0 and f(4) = 20.
✏️ Quick check · 3 of 8
Rolle’s theorem
f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b). Then:
The average slope is 0.
✍️ Type your answer · 4 of 8
Find the limit as x→0 of (eˣ − 1)/x.
What is the limit?
Enter a whole number or a fraction such as -3/4.
It has the form 0/0.
✍️ Type your answer · 5 of 8
Find the limit as x→0 of (1 − cos x)/x².
What is the limit?
Enter a whole number or a fraction such as -3/4.
After one step it is still 0/0.
✍️ Type your answer · 6 of 8
Find the limit as x→1 of (ln x)/(x − 1).
What is the limit?
Enter a whole number or a fraction such as -3/4.
ln 1 = 0, so it has the form 0/0.
✍️ Type your answer · 7 of 8
Find the limit as x→0⁺ of x ln x.
What is the limit?
Enter a whole number or a fraction such as -3/4.
Rewrite 0 · ∞ as a fraction.
✏️ Quick check · 8 of 8
Misuse
Why is L’Hôpital’s rule wrong for (x + 1)/(x + 2) as x→1?
What does direct substitution give?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The Mean Value Theorem requires f to be:
Question 2 of 5
If f′(x) = 0 on an interval, then f is:
Question 3 of 5
L’Hôpital’s rule differentiates:
Question 4 of 5
The limit of (1 + 1/x)ˣ as x→∞ is:
Question 5 of 5
The limit of x²/eˣ as x→∞ is:
Lesson 10 · Integrals
🧱 Antiderivatives & the Definite Integral
Integration has two faces: undoing a derivative and adding up infinitely many small pieces. This lesson introduces each one; Lesson 11 shows that they are the same thing.
↩️ Antiderivatives
F is an antiderivative of f if F′ = f. All antiderivatives differ by a constant, so we write ∫ f(x) dx = F(x) + C.
∫ xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1)
∫ 1/x dx = ln|x| + C
∫ eˣ dx = eˣ + C
∫ sin x dx = −cos x + C
∫ cos x dx = sin x + C
∫ sec²x dx = tan x + C
An initial condition fixes C. If F′(x) = 2x and F(0) = 5, then F(x) = x² + 5.
🖼️ See it
A right-endpoint sum with four strips overestimates the area under y = x² on [0, 2]: 3.75 against the exact 8/3.
🖼️ See it
The integral of sin x from 0 to 2π is 0: the region above the axis cancels the region below.
🧱 Riemann sums
To find the area under a curve on [a, b], split it into n strips of width Δx = (b − a)/n, and add up rectangles: Σ f(xᵢ) Δx. As n→∞, the sum approaches the definite integral:
∫ from a to b of f(x) dx = lim as n→∞ of Σ f(xᵢ) Δx
f(x) = x² on [0, 2], with n = 4 and right endpoints:
Δx = 0.5
0.5 × (0.25 + 1 + 2.25 + 4) = 3.75
The exact value is 8/3 ≈ 2.67
Right endpoints overestimate an increasing function; left endpoints underestimate it.
📋 Properties of the definite integral
∫ from a to a of f = 0
∫ from b to a of f = −∫ from a to b of f
∫ from a to b + ∫ from b to c = ∫ from a to c
∫ (cf + g) = c ∫ f + ∫ g
Signed area. Area below the x-axis counts as negative. For total area, split the integral where f changes sign.
Common mistake. A definite integral is a number, so it has no + C. An indefinite integral is a family of functions, so it must.
✍️ Type your answer · 1 of 8
Approximate the area under f(x) = x² on [0, 2] with 2 strips of width 1 and right endpoints.
What is the sum?
Enter a whole number or a fraction such as -3/4.
Use f(1) and f(2).
✍️ Type your answer · 2 of 8
Now use the same 2 strips with left endpoints.
What is the sum?
Enter a whole number or a fraction such as -3/4.
Use f(0) and f(1).
✍️ Type your answer · 3 of 8
The region under f(x) = 2x from x = 0 to x = 3 is a triangle.
What is the exact value of the integral?
Enter a whole number or a fraction such as -3/4.
Base 3, height f(3) = 6.
✏️ Quick check · 4 of 8
Antiderivatives
∫ 3x² dx equals:
Raise the power, divide by the new power.
✏️ Quick check · 5 of 8
Antiderivatives
∫ cos x dx equals:
Check by differentiating your answer.
✍️ Type your answer · 6 of 8
F′(x) = 2x and F(0) = 5.
What is F(3)?
Enter a whole number or a fraction such as -3/4.
Use the initial condition to find C.
✍️ Type your answer · 7 of 8
A particle has velocity v(t) = 6t and starts at s(0) = 1.
What is its position s(2)?
Enter a whole number or a fraction such as -3/4.
Position is the antiderivative of velocity.
✍️ Type your answer · 8 of 8
The integral of f from 0 to 3 is 7.
What is the integral of f from 3 to 0?
Enter a whole number or a fraction such as -3/4.
Swap the limits.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
∫ 1/x dx equals:
Question 2 of 5
∫ sin x dx equals:
Question 3 of 5
The integral from 2 to 2 of any function is:
Question 4 of 5
For an increasing function, a left-endpoint sum:
Question 5 of 5
Why does an indefinite integral need + C?
Lesson 11 · Integrals
🏛️ The Fundamental Theorem of Calculus
Derivatives and areas look unrelated. The Fundamental Theorem shows they are inverse processes, and turns a limit of infinite sums into a one-line calculation.
🏛️ Part 2: evaluating integrals
If F is any antiderivative of a continuous function f, then:
∫ from a to b of f(x) dx = F(b) − F(a)
∫ from 0 to 2 of x² dx = [x³/3] from 0 to 2 = 8/3 − 0 = 8/3
∫ from 0 to π of sin x dx = [−cos x] from 0 to π = 1 − (−1) = 2
∫ from 1 to 4 of √x dx = [(2/3)x^(3/2)] from 1 to 4 = 16/3 − 2/3 = 14/3
The constant cancels in the subtraction, so you may leave out + C. This confirms the exact value of 8/3 from Lesson 10.
🖼️ See it
The shaded area from 0 to 2 is F(2) − F(0) = 8/3, found without adding any rectangles.
🖼️ See it
⚙️ Part 1: differentiating integrals
If f is continuous and G(x) = ∫ from a to x of f(t) dt, then G′(x) = f(x). The rate at which area accumulates equals the height of the curve at that point.
d/dx ∫ from 1 to x of t³ dt = x³
If the upper limit is a function, use the chain rule:
d/dx ∫ from 0 to x² of eᵗ dt = e^(x²) · 2x
Common mistake. With an upper limit of x², forgetting the factor 2x.
🚗 Net change
Part 2 can be read as: the integral of a rate gives the total change, ∫ F′ = F(b) − F(a). It applies to anything with a rate.
Velocity v(t) = 3t² m/s from t = 0 to t = 2:
Displacement = [t³] from 0 to 2 = 8 m
For total distance travelled, integrate |v(t)|,
splitting wherever v changes sign.
✍️ Type your answer · 1 of 8
Evaluate the integral of x² from 0 to 2.
What is its value?
Enter a whole number or a fraction such as -3/4.
The antiderivative of x² is x³/3.
✍️ Type your answer · 2 of 8
Evaluate the integral of 2x from 1 to 3.
What is its value?
Enter a whole number or a fraction such as -3/4.
The antiderivative of 2x is x².
✍️ Type your answer · 3 of 8
Evaluate the integral of 3x² + 1 from 0 to 1.
What is its value?
Enter a whole number or a fraction such as -3/4.
Integrate each term.
✍️ Type your answer · 4 of 8
Evaluate the integral of sin x from 0 to π.
What is its value?
Enter a whole number or a fraction such as -3/4.
cos π = −1 and cos 0 = 1.
✍️ Type your answer · 5 of 8
Evaluate the integral of √x from 1 to 4.
What is its value?
Enter a whole number or a fraction such as -3/4.
Write √x as x^(1/2).
✏️ Quick check · 6 of 8
Part 1
d/dx of the integral from 0 to x of cos t dt is:
No integrating is needed.
✏️ Quick check · 7 of 8
Part 1 with the chain rule
d/dx of the integral from 0 to x² of eᵗ dt is:
The upper limit is a function of x.
✍️ Type your answer · 8 of 8
G(x) is the integral of t³ from 1 to x.
What is G′(2)?
Enter a whole number or a fraction such as -3/4.
Use Part 1, then substitute.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The Fundamental Theorem links:
Question 2 of 5
The integral of a velocity function gives:
Question 3 of 5
Why can + C be dropped in a definite integral?
Question 4 of 5
The integral of eˣ from 0 to 1 is:
Question 5 of 5
The integral of 1/x from 1 to e is:
Lesson 12 · Integrals
🔄 Integration by Substitution
There is no product rule or chain rule for integrals that always works. Substitution is the chain rule run backwards, and it is the first technique to try.
🔄 The method
Look for a function and its derivative in the integrand. Set u to the inner function, and find du in terms of dx.
∫ 2x cos(x²) dx
u = x², du = 2x dx
= ∫ cos u du = sin u + C = sin(x²) + C
Check by differentiating: cos(x²) · 2x ✓
Always finish an indefinite integral in terms of x, and check by differentiating.
🖼️ See it
🖼️ See it
🧪 Fixing constants
The derivative only has to match up to a constant factor, since you can divide it out.
∫ (2x + 1)³ dx
u = 2x + 1, du = 2 dx, so dx = du/2
= (1/2) ∫ u³ du = u⁴/8 + C = (2x + 1)⁴/8 + C
∫ tan x dx = ∫ sin x / cos x dx
u = cos x, du = −sin x dx
= −∫ du/u = −ln|cos x| + C
∫ 1/(x + 3) dx = ln|x + 3| + C
Common mistake. A factor of x cannot be treated as a constant. ∫ e^(x²) dx has no 2x in it, so this substitution fails, and it has no elementary antiderivative at all.
🎚️ Definite integrals: change the limits
Convert the limits to u-values at the same time as the integrand. Then you never need to return to x.
∫ from 0 to 1 of 2x(x² + 1)³ dx
u = x² + 1, du = 2x dx
x = 0 gives u = 1. x = 1 gives u = 2.
= ∫ from 1 to 2 of u³ du = [u⁴/4] from 1 to 2 = 4 − 1/4 = 15/4
✏️ Quick check · 1 of 8
Substitution
∫ 2x cos(x²) dx equals:
The 2x is the derivative of x².
✏️ Quick check · 2 of 8
Logs
∫ 1/(x + 3) dx equals:
The derivative of the bottom is 1.
✍️ Type your answer · 3 of 8
Evaluate the integral of 2x(x² + 1)³ from 0 to 1.
What is its value?
Enter a whole number or a fraction such as -3/4.
Change the limits to u-values.
✍️ Type your answer · 4 of 8
Evaluate the integral of sin x cos x from 0 to π/2.
What is its value?
Enter a whole number or a fraction such as -3/4.
Let u = sin x.
✍️ Type your answer · 5 of 8
Evaluate the integral of x/√(x² + 16) from 0 to 3.
What is its value?
Enter a whole number or a fraction such as -3/4.
Let u = x² + 16, so x dx = du/2.
✍️ Type your answer · 6 of 8
Evaluate the integral of (2x + 1)³ from 0 to 1.
What is its value?
Enter a whole number or a fraction such as -3/4.
Let u = 2x + 1. Do not forget the factor of 1/2.
✏️ Quick check · 7 of 8
Limits
For a definite integral, after substituting you should:
The limits are values of the variable you integrate in.
✏️ Quick check · 8 of 8
Trig
∫ tan x dx equals:
Write tan x as sin x / cos x.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Substitution reverses the:
Question 2 of 5
For ∫ f′(x)/f(x) dx, the answer is:
Question 3 of 5
For ∫ x e^(x²) dx, the best choice is:
Question 4 of 5
∫ x e^(x²) dx equals:
Question 5 of 5
∫ (3x − 2)⁴ dx equals:
Lesson 13 · Integrals
🏺 Areas, Averages & Volumes
The definite integral adds up thin slices of anything. Pick the right slice and the same idea measures the area between curves, the average of a function and the volume of a solid.
🗺️ Area between curves
If f ≥ g on [a, b], the area between them is ∫ from a to b of (f − g) dx: top minus bottom. Find the limits where the curves meet.
y = x and y = x²
They meet where x² = x: x = 0 and x = 1
On [0, 1], x is above x²
Area = ∫ from 0 to 1 of (x − x²) dx = 1/2 − 1/3 = 1/6
If the curves cross inside the interval, split the integral there. Area is always positive, so the top curve changes.
🖼️ See it
The shaded region between y = x and y = x² has area 1/6: top minus bottom.
🖼️ See it
⚖️ Average value
The average value of f on [a, b] is the height of the rectangle with the same area:
f_avg = (1/(b − a)) ∫ from a to b of f(x) dx
f(x) = x² on [0, 3]
f_avg = (1/3)(9) = 3
The Mean Value Theorem for Integrals says a continuous function actually reaches its average value somewhere on the interval.
🏺 Volumes by disks
Rotate y = f(x) around the x-axis. Each thin slice is a disk of radius f(x), so the volume is the sum of the disks:
V = π ∫ from a to b of [f(x)]² dx
y = √x on [0, 2]
V = π ∫ from 0 to 2 of x dx = π[x²/2] = 2π
Cone: y = x on [0, 3]
V = π ∫ from 0 to 3 of x² dx = 9π
This agrees with (1/3)πr²h = (1/3)π(9)(3) = 9π
With a hole in the solid, use washers: π ∫ (R² − r²) dx, with outer radius R and inner radius r.
Common mistake. Forgetting to square the radius: the area of a disk is πr².
✍️ Type your answer · 1 of 8
The region between y = x and y = x².
What is its area?
Enter a whole number or a fraction such as -3/4.
They meet at x = 0 and x = 1.
✍️ Type your answer · 2 of 8
The region between y = 4 − x² and the x-axis.
What is its area?
Enter a whole number or a fraction such as -3/4.
The parabola meets the axis at x = ±2.
✍️ Type your answer · 3 of 8
The region between y = x + 2 and y = x².
What is its area?
Enter a whole number or a fraction such as -3/4.
Solve x² = x + 2 for the limits.
✍️ Type your answer · 4 of 8
f(x) = x² on [0, 3].
What is the average value of f?
Enter a whole number or a fraction such as -3/4.
Integrate, then divide by the width.
✍️ Type your answer · 5 of 8
f(x) = 2x + 1 on [0, 4].
What is the average value of f?
Enter a whole number or a fraction such as -3/4.
Integrate, then divide by 4.
✏️ Quick check · 6 of 8
Volumes
Rotating y = f(x) about the x-axis gives a volume of:
What is the area of each disk?
✍️ Type your answer · 7 of 8
Rotate y = √x on [0, 2] about the x-axis.
The volume is what multiple of π?
Enter the number that multiplies π. For 5π, enter 5.
Squaring √x gives x.
✍️ Type your answer · 8 of 8
Rotate y = x on [0, 3] about the x-axis, which makes a cone.
The volume is what multiple of π?
Enter the number that multiplies π. For 5π, enter 5.
Square the radius, then integrate.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The area between two curves is the integral of:
Question 2 of 5
The limits for the area between two curves usually come from:
Question 3 of 5
The average value of f(x) = 6 on [1, 5] is:
Question 4 of 5
The washer method is used when the solid has:
Question 5 of 5
Rotating y = 2 on [0, 5] about the x-axis gives a:
Final Exam
🏆 20-Question Final Exam
Questions from every lesson in this course. Pick an answer for each one, then hit Check My Answers.
0 / 20
💪 Keep practicing — go back through the lessons and retake the quiz!
Q1 · Limits
The limit of (x² − 25)/(x − 5) as x→5 is:
Q2 · Limits
The limit of sin(4x)/x as x→0 is:
Q3 · Continuity & Limits at Infinity
The limit of (6x² + x)/(3x² − 2) as x→∞ is:
Q4 · Continuity & Limits at Infinity
The IVT guarantees a root when a continuous function:
Q5 · The Derivative
f′(a) is geometrically the:
Q6 · Differentiation Rules
The derivative of x³ sin x is:
Q7 · Differentiation Rules
The derivative of (x + 1)/x, for x ≠ 0, is:
Q8 · Chain Rule & Implicit Differentiation
The derivative of sin²x is:
Q9 · Chain Rule & Implicit Differentiation
For x² + y² = 13, dy/dx at (2, 3) is:
Q10 · Related Rates & Linear Approximation
A circle’s radius grows at 1 cm/s. When r = 4, dA/dt in cm²/s is:
Q11 · Curve Sketching
If f′(c) = 0 and f″(c) > 0, then c is a:
Q12 · Curve Sketching
f(x) = x² − 4x on [0, 5] has an absolute maximum of:
Q13 · Optimization
Two positive numbers add to 12. Their greatest product is:
Q14 · Mean Value Theorem & L’Hôpital’s Rule
For f(x) = x² on [0, 4], the Mean Value Theorem gives c =
Q15 · Mean Value Theorem & L’Hôpital’s Rule
The limit of (sin x − x)/x³ as x→0 is:
Q16 · Antiderivatives & the Definite Integral
∫ (4x³ + 2) dx equals:
Q17 · The Fundamental Theorem of Calculus
The integral of 3x² from 1 to 2 is:
Q18 · The Fundamental Theorem of Calculus
d/dx of the integral from 2 to x of (t² + 1) dt is:
Q19 · Integration by Substitution
∫ 3x²(x³ + 1)⁴ dx equals:
Q20 · Areas, Averages & Volumes
Rotating y = x on [0, 2] about the x-axis gives a volume of: