Turning points have flat tangents.
Climb a hill: rising before the top, falling after — so at the very top the slope passes through zero. The same holds at the bottom of a valley. That is the whole strategy: turning points hide where f′(x) = 0.
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Maxima and Minima with Derivatives
At the top of every peak and the bottom of every valley, the tangent goes flat. Set f′(x) = 0 and the interesting points solve themselves.
Read the graph
At which x does the local maximum occur?
The peak sits at (−1, 2), so the local maximum occurs at x = −1. A is the valley — the local minimum. B is where the curve crosses the axis (a root, not a peak), and at 0 the curve is falling straight through, not turning.
Set the derivative to zero and solve.
f(x) = x³ − 3x → f′(x) = 3x² − 3
power rule, term by term
3x² − 3 = 0 → x² = 1
the flat-tangent condition
x = −1 or x = 1
two critical points — the graph shows both
A known critical number pins down a missing coefficient. If f(x) = x³ + bx² + 24x has a critical number at x = 2, then f′(2) = 12 + 4b + 24 = 0, so b = −9. Now f′(x) = 3x² − 18x + 24 = 3(x − 2)(x − 4), and the other critical number is x = 4.
Which is the peak, which is the valley?
Check the heights, or watch the sign of f′ change as you pass through.
f(−1) = −1 + 3 = 2
the higher one — local maximum
f(1) = 1 − 3 = −2
the lower one — local minimum
| f′ changes | The point is a… |
|---|---|
| + to − | peak — rising, then falling |
| − to + | valley — falling, then rising |
Trap: answering with the wrong number
Asked for the local maximum value of f(x) = x³ − 3x, a student solves f′(x) = 0, gets x = −1, and writes “−1”. That is where the peak is; its height is f(−1) = 2. A second slip: taking only the positive root of x² = 1 gives x = 1, and f(1) = −2 gets reported as the maximum — but that point is the valley. Keep both roots, then read which one the question asks for: the x, or f(x).
For parabolas, this is a formula you already know.
y = ax² + bx + c → y′ = 2ax + b = 0
flat-tangent condition
x = −b2a
the vertex formula is calculus in disguise
Every "maximum height" and "minimum cost" question about a quadratic is this one line. The derivative just extends the same idea to curves the vertex formula cannot touch.
Word problems: one formula in one letter, then f′ = 0.
240 m of fencing makes a rectangular field split into two pens by one extra fence parallel to the ends. What is the largest total area?
ends and divider: 3 fences of length x
draw it and count the pieces
3x + 2L = 240 → L = 120 − 1.5x
the constraint removes the second letter
A = x(120 − 1.5x) = 120x − 1.5x²
the thing to maximise, in x only
A′ = 120 − 3x = 0 → x = 40, L = 60
flat tangent
A = 40 × 60 = 2400 m²
check: 3(40) + 2(60) = 240 ✓
The best shape is not always a square
A square of side 48 uses 3(48) + 2(48) = 240 m but encloses only 2304 m². Extra fences (a divider) or missing ones (a wall or river side) shift the best shape. Write the fence equation from the picture, never from habit.
Two numbers with a fixed sum. Two positive numbers add to 15; make the first times the square of the second as large as possible. With the second = y: P = (15 − y)y² = 15y² − y³, P′ = 30y − 3y² = 3y(10 − y) = 0, so y = 10, the first is 5, and P = 5 × 100 = 500.
A rectangle under a curve. Its base is on the x-axis and its top corners are on y = 27 − x². With corners at x and −x, the width is 2x and the height 27 − x²: A = 2x(27 − x²) = 54x − 2x³, A′ = 54 − 6x² = 0, so x = 3 and A = 6 × 18 = 108.
Variations: a flat spot that is not a turn, a closed interval, a 1x term.
f′ = 0 but no turn. Classify the critical point of f(x) = x³ − 6x² + 12x.
f′(x) = 3x² − 12x + 12 = 3(x − 2)²
a perfect square
f′ = 0 only at x = 2
one critical number
f′(1) = 3, f′(3) = 3
+ before, + after: no change of sign
neither — a shelf at (2, 8)
f(2) = 8 − 24 + 24 = 8
What is different: a flat tangent is necessary for a turn but not enough. The curve pauses at x = 2 and keeps rising.
A closed interval. Find the largest and smallest values of f(x) = x³ − 3x² for −1 ≤ x ≤ 4.
f′ = 3x² − 6x = 3x(x − 2) = 0 → x = 0, 2
both inside the interval
f(0) = 0, f(2) = 8 − 12 = −4
the turning points
f(−1) = −1 − 3 = −4, f(4) = 64 − 48 = 16
the two ends
largest 16 (at x = 4), smallest −4
−4 occurs twice: x = −1 and x = 2
What is different: the biggest value sits at an END, where f′ is not 0. On a closed interval, always test the endpoints too.
The local maximum is not the largest value
f(0) = 0 is a peak, so it is tempting to answer 0. But the curve climbs past it to 16 at the right end. “Local” means only among nearby points.
A 1x term, positive x only. Find the minimum of f(x) = x + 9x for x > 0.
f′(x) = 1 − 9x² = 0
9x−1 → −9x−2
x² = 9 → x = 3 (not −3)
−3 breaks x > 0
f(3) = 3 + 3 = 6
check: f(2) = 6.5, f(4) = 6.25, both bigger ✓
What is different: not a polynomial, and the given condition throws one root away before we even classify.
Throw out critical points the problem cannot use.
An open box is made from a 10 cm by 16 cm sheet by cutting squares of side x from the corners. Which x gives the largest volume?
V = x(10 − 2x)(16 − 2x) = 4x³ − 52x² + 160x
height × width × length, expanded
V′ = 12x² − 104x + 160 = 4(3x − 20)(x − 2)
power rule, then factor
x = 2 or x = 203
two candidates
203 > 5: width 10 − 2x would be negative
the domain is 0 < x < 5
x = 2 cm, V = 2 × 6 × 12 = 144 cm³
the only usable one
On the exam
Expect a minimum or maximum value of a polynomial, or a story — fence, box, two numbers with a fixed sum. Build one formula in one letter, set its derivative to zero, reject impossible x, and answer what was asked: the x, the side lengths, or the area itself. Plugging a neighbouring x in (it must give a worse value) is a fast check.
Check the sign of f′ on both sides of each point.
f′(−2)
3(4) − 3 = +9
rising into x = −1…
f′(0)
−3
…falling after it: + to −, a peak ✓
f′(2)
+9
rising after x = 1: − to +, a valley ✓
Now you try
Practice problem
The problem
At which x does f(x) = x² − 4x + 1 reach its minimum?
f′(x) = 2x − 4 = 0 gives x = 2 — and since the parabola opens upward, that flat point is the minimum. A drops the sign in −b2a (b is −4, so −b is +4). It agrees with the vertex formula: x = 42 = 2.