The derivative shows where a function rises, where it falls, and where it turns around.
Where the slope becomes 0 is a candidate for a peak, a local maximum, or a valley, a local minimum, and the sign of the second derivative tells them apart.
The same idea solves optimization problems, finding the largest or smallest possible value of something like an area or a cost.
In the first picture, changing a moves the maximum and minimum; in the second, changing the width shows that a rectangle with perimeter 20 has the greatest area as the 5 by 5 square.
Increasing, decreasing and extrema
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f(x) = x³ − ax (gold) and its derivative f'(x) = 3x² − a (blue dashes); red points are the local maximum and minimum, the green point the inflection point
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👀 What to read in the picture
①The gold curve is f(x); the blue dashed curve is f'(x)
②f increases where f'(x) > 0 and decreases where f'(x) < 0
③The red points sit at x = ±√(a/3), where f'(x) = 0
④When a = 0, f' ≥ 0 everywhere, so there is no extremum
Identifying extrema
First-derivative test
f'(c) = 0 and f' sign + → − : max; − → + : min
The way the sign of f' changes separates maxima from minima
Second-derivative test
f'(c) = 0, f''(c) > 0 → min; f''(c) < 0 → max
The sign of f''(c) shows which way the curve bends, which settles the extremum
🏔️ Peaks and valleys
①A local maximum is a peak, where the curve is concave down (f'' < 0)
②A local minimum is a valley, where the curve is concave up (f'' > 0)
③If f''(c) = 0 the test says nothing, so check the sign of f' instead
Optimization
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A rectangle with perimeter 20 (left) and the graph of its area A(x) = x(10 − x) against the width x (right)
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📐 Steps for optimization
①Write the quantity as a function: the height is 10 − x, so A(x) = x(10 − x)
②Differentiate and set it to 0: A'(x) = 10 − 2x = 0 gives x = 5
③Test it: A''(5) = −2 < 0, so it is a maximum
④The maximum is A(5) = 25, and the rectangle is then a square
Tangent lines and approximation
Equation of the tangent
y − f(a) = f'(a)(x − a)
The tangent at x = a has slope f'(a) and passes through (a, f(a))
Linear approximation
f(x) ≈ f(a) + f'(a)(x − a) (x ≈ a)
Near a, the function can be estimated by its tangent line
💡 Estimating with a tangent
①Estimating √4.01 with the tangent at √4 gives 2 + (1/4) × 0.01 = 2.0025
②The tangent is the straight line closest to the function near that point
③Going further leads to approximations that also use squared and cubed terms
Worked examples
Example 1
Find the local extrema of f(x) = x3 − 3x.
1
Find the points where f'(x) = 0 as candidates for extrema.
f'(x) = 3x2 − 3 = 0 ⇒ x = ±1
2
Use the sign of the second derivative f''(x) = 6x.
f''(−1) < 0 ⇒ local max 2, f''(1) > 0 ⇒ local min −2
▸ local max 2 (x = −1), local min −2 (x = 1)
f''(−1) = −6 < 0 makes x = −1 a peak, and f''(1) = 6 > 0 makes x = 1 a valley. Setting a = 3 in the first picture gives this very curve.
Example 2
Find the tangent line to y = x3 − 2x at the point (1, −1).
1
Find the slope of the tangent, f'(1).
f'(x) = 3x2 − 2, f'(1) = 1
2
Put it into y − f(a) = f'(a)(x − a).
y − (−1) = 1·(x − 1) ⇒ y = x − 2
▸ y = x − 2
A tangent line needs only the slope f'(a) and the point of contact (a, f(a)).
Summary
Application Core
f' = 0 → extremum candidate; f'' > 0 → min; f'' < 0 → max
Find candidates with the derivative, then test them with the second derivative or a change of sign
CSAT-style problem, adapted
On 0 ≤ x ≤ 3, what is the maximum of f(x) = x3 − 3x2 + 4?
①0
②2
③4
④6
⑤8
▸ ③ 4
1
Find the points where f'(x) = 0.
f'(x) = 3x(x − 2) = 0 ⇒ x = 0, 2
2
Compare the function values at the ends of the closed interval and at the candidates.
f(0)=4, f(2)=0, f(3)=4 ⇒ maximum 4
🎯 Exam Points
①Sign chart: record the sign of f' on each interval
②Decide extrema by the sign change of f' or the sign of f''
③Optimization: quantity as a function → differentiate → zeros → test
④The tangent line is y − f(a) = f'(a)(x − a)
⑤An inflection point has f'' = 0 with f'' changing sign