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Increasing/Decreasing & 1st Derivative Test

Unit 5 · Analytical Applications of Differentiation

Key Ideas

The sign of f′ tells you whether f is rising or falling. At a critical point where f′ changes sign, f has a local extremum.

Increasing / Decreasing

f′(x) > 0 on (a, b) →
f is increasing on (a,b)f \text{ is increasing on } (a,b)
f′(x) < 0 on (a, b) →
f is decreasing on (a,b)f \text{ is decreasing on } (a,b)

First Derivative Test — at critical point x = c

f′ changes − → + →
local minimum at c\text{local minimum at } c
f′ changes + → − →
local maximum at c\text{local maximum at } c
f′ no sign change →
neither (e.g. inflection point)\text{neither (e.g. inflection point)}

Candidates Test (Closed Interval [a, b])

Evaluate f at:
all critical points in (a,b) and both endpoints a,b\text{all critical points in } (a,b) \text{ and both endpoints } a,\, b
Then:
largest value=abs. max,smallest value=abs. min\text{largest value} = \text{abs. max}, \quad \text{smallest value} = \text{abs. min}
Type 1

Increasing / Decreasing Intervals

Find the sign of f′ on each interval between critical points to determine where f is increasing (f′ > 0) or decreasing (f′ < 0).

Example 1

Find all intervals on which f is increasing and decreasing.

f(x)=x33x29x+2f(x) = x^3 - 3x^2 - 9x + 2

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Type 2

Relative Extrema

At each critical point, check if f′ changes sign. Negative → positive means local minimum; positive → negative means local maximum.

Example 2

Find all relative maxima and minima of f.

f(x)=x33x29x+2f(x) = x^3 - 3x^2 - 9x + 2

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Type 3

Absolute Extrema (Candidates Test)

On a closed interval [a, b], the absolute max and min must occur at a critical point or at an endpoint. Evaluate f at all candidates and compare.

Example 3

Find the absolute maximum and minimum values of f on [−2, 4].

f(x)=x33x29x+2,[2,4]f(x) = x^3 - 3x^2 - 9x + 2, \quad [-2,\, 4]

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1st Derivative Test | AP Calculus AB — Mathfolis