Increasing, Decreasing, and Local Extrema

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\(f' > 0\) increasing, \(f' < 0\) decreasing

sign change decides a local max or min

What to Remember

The sign of f' is the direction of f

\(f'(x) > 0\) on an interval means \(f\) is increasing there. \(f'(x) < 0\) means \(f\) is decreasing there.

The first derivative test

At a critical point, a change from positive to negative is a local maximum. A change from negative to positive is a local minimum.

If the sign does not change, the critical point is not a local extremum.

Test intervals, not just the zeros

The zeros of \(f'\) and any x-values where \(f'\) is undefined split the domain. Pick one test number in each open piece.

A discontinuity splits the domain

If \(f\) is undefined at a number, that number is not a local extremum. State the increasing and decreasing intervals on each side of it.

Level 1

Easy

The derivative is already factored, or it is a simple quadratic.

Example

For \(f(x) = x^3 - 3x\), find the intervals where \(f\) is increasing and where it is decreasing.

\(f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)\)

Critical points: \(x = -1\) and \(x = 1\).

Test \(x = -2\): \(f' > 0\).

Test \(x = 0\): \(f' < 0\).

Test \(x = 2\): \(f' > 0\).

Increasing on \((-\infty, -1)\) and \((1, \infty)\). Decreasing on \((-1, 1)\).

Answer: increasing on \((-\infty,-1)\) and \((1,\infty)\); decreasing on \((-1,1)\)

For \(f(x) = x^3 - 3x\), classify the critical points with the first derivative test.

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At \(x = -1\), \(f'\) changes from positive to negative. Local maximum.

\(f(-1) = -1 + 3 = 2\)

At \(x = 1\), \(f'\) changes from negative to positive. Local minimum.

\(f(1) = 1 - 3 = -2\)

Answer: local max \((-1, 2)\); local min \((1, -2)\)

\(f'(x) = x - 2\). On which intervals is \(f\) increasing, and what kind of extremum is at \(x = 2\)?

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\(f'(x) > 0\) for \(x > 2\), and \(f'(x) < 0\) for \(x < 2\).

The sign changes from negative to positive, so \(x = 2\) is a local minimum.

Answer: decreasing on \((-\infty,2)\); increasing on \((2,\infty)\); local min at 2

\(f'(x) = -(x - 1)^2\). Describe \(f\) at the critical point \(x = 1\).

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\(f'(x) \le 0\) everywhere, and \(f'(x) = 0\) only at \(x = 1\).

The sign does not change.

\(x = 1\) is not a local extremum. f is decreasing on \((-\infty, \infty)\).

Answer: not a local extremum

For \(f(x) = x^2 - 4x\), find the local minimum value.

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\(f'(x) = 2x - 4\)

\(x = 2\)

\(f'\) changes from negative to positive.

\(f(2) = 4 - 8 = -4\)

Answer: local min value \(-4\)

If \(f'\) changes from positive to negative at \(x = 4\), what does \(f\) have there?

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A change from positive to negative is a local maximum.

Answer: a local maximum

Level 2

Medium

Three critical points, a restricted domain, or an exponential.

Example

For \(f(x) = x^4 - 8x^2\), find the local maxima and local minima.

\(f'(x) = 4x^3 - 16x = 4x(x - 2)(x + 2)\)

Critical points: \(x = -2, 0, 2\).

Signs of \(f'\): negative, positive, negative, positive.

Local minima at \(x = -2\) and \(x = 2\). Local maximum at \(x = 0\).

\(f(-2) = f(2) = -16\) and \(f(0) = 0\)

Answer: local min value \(-16\); local max value 0

For \(f(x) = x + \dfrac{4}{x}\) with \(x > 0\), find the local minimum value.

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\(f'(x) = 1 - \dfrac{4}{x^2}\)

\(x = 2\)

On \((0, 2)\), \(f' < 0\). On \((2, \infty)\), \(f' > 0\).

\(f(2) = 2 + 2 = 4\)

Answer: local min value 4

For \(f(x) = x e^{x}\), classify the critical point.

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\(f'(x) = e^{x}(x + 1)\)

\(x = -1\)

\(e^{x} > 0\), so \(f'\) has the sign of \((x + 1)\).

Negative to positive: local minimum.

\(f(-1) = -\dfrac{1}{e}\)

Answer: local min \(\left(-1, -\dfrac{1}{e}\right)\)

\(f'(x) = (x - 1)^2(x + 2)\). Where is \(f\) decreasing, and where is the local minimum?

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The factor \((x - 1)^2\) is never negative.

\(f'(x) < 0\) when \(x + 2 < 0\), so on \((-\infty, -2)\).

\(f'(x) > 0\) on \((-2, 1)\) and \((1, \infty)\), except \(f'(1) = 0\).

Local minimum at \(x = -2\). The critical point \(x = 1\) does not change sign.

Answer: decreasing on \((-\infty,-2)\); local min at \(x = -2\)

Which x-value is a local maximum of \(f(x) = x^4 - 8x^2\)?

A. \(0\)

B. \(2\)

C. \(-2\)

D. \(4\)

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\(f'\) changes from positive to negative only at \(x = 0\).

Answer: A. \(0\)

For \(f(x) = x^3 - 3x\), give the local maximum as a point.

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Sign change from positive to negative at \(x = -1\).

\(f(-1) = 2\)

Answer: \((-1, 2)\)

Level 3

Hard

A rational function, a trigonometric function, and a written justification.

Example

For \(f(x) = \dfrac{x^2}{x - 2}\), find the local extrema.

\(f'(x) = \dfrac{x(x - 4)}{(x - 2)^2}\)

The denominator is positive wherever it is defined.

\(f' > 0\) on \((-\infty, 0)\) and \((4, \infty)\).

\(f' < 0\) on \((0, 2)\) and \((2, 4)\).

Local maximum at \(x = 0\): \(f(0) = 0\).

Local minimum at \(x = 4\): \(f(4) = 8\).

Answer: local max \((0, 0)\); local min \((4, 8)\)

For \(f(x) = \sin x + \cos x\) on \((0, 2\pi)\), find the local extrema.

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\(f'(x) = \cos x - \sin x\)

\(\tan x = 1\)

In \((0, 2\pi)\), \(x = \dfrac{\pi}{4}\) and \(x = \dfrac{5\pi}{4}\).

Test \(x = 0\): \(f' > 0\). Test \(x = \pi/2\): \(f' < 0\). Local max at \(\pi/4\).

Test \(x = 3\pi/2\): \(f' > 0\). Local min at \(5\pi/4\).

\(f\!\bigl(\tfrac{\pi}{4}\bigr) = \sqrt{2}\) and \(f\!\bigl(\tfrac{5\pi}{4}\bigr) = -\sqrt{2}\)

Answer: local max \((\pi/4, \sqrt{2})\); local min \((5\pi/4, -\sqrt{2})\)

Justify the local maximum of \(f(x) = x^3 - 3x\) at \(x = -1\) in one chain of reasons a reader can check.

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\(f'(x) = 3(x - 1)(x + 1)\)

\(f'(x) > 0\) on \((-\infty, -1)\) and \(f'(x) < 0\) on \((-1, 1)\).

By the first derivative test, \(f\) has a local maximum at \(x = -1\).

\(f(-1) = 2\)

Answer: local max at \((-1, 2)\)

For \(f(x) = \dfrac{x^2}{x - 2}\), on which intervals is \(f\) decreasing?

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\(f'(x) = \dfrac{x(x - 4)}{(x - 2)^2}\)

The quotient is negative on \((0, 2)\) and on \((2, 4)\).

Those intervals stay separate because \(x = 2\) is not in the domain.

Answer: \((0, 2)\) and \((2, 4)\)

A function has \(f'(x) = (x + 1)(x - 4)\). Which statement is true?

A. Local max at \(x = -1\) and local min at \(x = 4\)

B. Local min at \(x = -1\) and local max at \(x = 4\)

C. Local max at both

D. No local extrema

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Signs of \(f'\): positive, negative, positive.

Positive to negative at \(x = -1\): local max.

Negative to positive at \(x = 4\): local min.

Answer: A. local max at \(-1\), local min at \(4\)

\(f'\) is continuous, \(f'(x) < 0\) for \(x < 0\), \(f'(0) = 0\), and \(f'(x) > 0\) for \(x > 0\). What does \(f\) have at \(x = 0\)?

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\(f'\) changes from negative to positive at \(x = 0\).

By the first derivative test, \(f\) has a local minimum at \(x = 0\).

Answer: a local minimum

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