Applications of Derivatives

Derivatives are a tool for spotting where a function rises, where it falls, and where it turns around. Where the slope hits zero sits a candidate for a peak or a valley, and the second derivative tells the two apart. The same idea drives optimization: finding the largest or smallest value of something like an area or a cost. Drag the coefficient a to watch the extrema shift, then change the width to see when the rectangle's area peaks.

Monotonicity & Extrema

Coefficient a3

👀 See It

①Gold curve is f(x); blue dashed is f'(x)

②f'(x) > 0 interval: f is increasing

③f'(x) < 0 interval: f is decreasing

④f'(x) = 0 (red): local max/min candidate

Extremum Tests

First Derivative Test

f'(c) = 0 and f' sign + → − : max; − → + : min

Decide max/min from sign change of f'

Second Derivative Test

f'(c) = 0, f''(c) > 0 → min; f''(c) < 0 → max

Use the sign of f'' (concavity) to decide

🏔️ Mountains and Valleys

①Local max = peak: concave down (f'' < 0)

②Local min = valley: concave up (f'' > 0)

③Inflection point: concavity flips (f'' = 0)

Optimization

Width x3

📐 Optimization Steps

①Objective: A(x) = x(10−x)

②Differentiate: A'(x) = 10 − 2x = 0 → x = 5

③Second test: A''(5) = −2 < 0 → max

④Maximum: A(5) = 25

Tangent & Linear Approximation

Tangent Line

y − f(a) = f'(a)(x − a)

At x = a: slope f'(a) through (a, f(a))

Linear Approximation

f(x) ≈ f(a) + f'(a)(x − a) (x ≈ a)

Approximate f near a by its tangent line

💡 Power of Approximation

①√(4.01) ≈ √4 + (1/2√4)(0.01) = 2.0025

②The tangent is 'the best linear approximation locally'

③This grows into Taylor series later

Worked Examples

Example 1

Find the local extrema of f(x) = x³ − 3x.

Find where f′(x) = 0 (the candidates for extrema).

f'(x) = 3x² − 3 = 0 ⇒ x = ±1

Classify using 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)

Use f′ = 0 to find candidates, then the sign of f″ to classify max (−) vs min (+).

Example 2

Find the tangent line to y = x³ − 2x at the point (1, −1).

Compute the slope f′(1).

f'(x) = 3x² − 2, f'(1) = 1

Substitute 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 (a, f(a)); use point-slope form.

Wrap-up

Application Core

f' = 0 → extremum candidate; f'' > 0 → min; f'' < 0 → max

First derivative for candidates, second for classification

CSAT-style

On 0 ≤ x ≤ 3, what is the maximum of f(x) = x³ − 3x² + 4?

①0

②2

③4

④6

⑤8

▸ ③ 4

Find the critical points where f′(x) = 0.

f'(x) = 3x(x − 2) = 0 ⇒ x = 0, 2

Compare values at the endpoints and critical points of the closed interval.

f(0)=4, f(2)=0, f(3)=4 ⇒ maximum 4

🎯 Exam Points

①Sign chart: organize sign of f' over intervals

②Choose 1st or 2nd derivative test

③Optimization: objective → diff → critical points → test

④Tangent: y − f(a) = f'(a)(x − a)

⑤Inflection: f'' = 0 with sign change

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Differentiation Methods

Second Derivative & Inflection

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