Derivative Test

Many applications of calculus require us to deduce facts about a function f from the information concerning its derivatives. Since f ‘ (x) represents the slope of the curve y = f(x) at the point (x, f(x)), it tells us the direction in which the curve proceeds at each point.

Increasing / Decreasing Test

1. If f(x0 > 0 on an interval, then f is increasing on that interval.
2. If f(x0 < 0 on an interval, then f is decreasing on that interval.

Example:

Find where the function f(x) = x³ – 12x + 1 is increasing and where it is decreasing.

Solution:

Step 1: Find the derivative of f

f ‘(x) = 3x² – 12 = 3(x² – 4) = 3(x –2) (x + 2)

Step 2: Set f ‘(x) = 0 to get the critical numbers

f ‘(x) = 3(x –2) (x + 2) = 0
x = 2, –2

Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f ‘(x) for each of the intervals. Use the increasing/decreasing test to determine whether f(x) is increasing or decreasing for each interval.

The First Derivative Test

Suppose that c is a critical number of a continuous function f.

1. If f ‘ changes from positive to negative at c, then f has a local maximum at c.
2. If f ‘ changes from negative to positive at c, then f has a local minimum at c.
3. If f ‘ does not change sign at c (f ‘ is positive at both sides of c or f ‘ is negative on both sides), then f has no local maximum or minimum at c.

Example:

Find the local maximum and minimum values of the function f(x) = x⁴ – 2x² + 3

Solution:

Step 1: Find the derivative of f

f ‘(x) = 4x³ – 4x = 4x(x² –1) = 4x(x –1)(x +1)

Step 2: Set f ‘(x) = 0 to get the critical numbers

f ‘(x) = 4x(x –1)(x +1) = 0
x = –1, 0, 1

Step 3: Set up intervals whose endpoints are the critical numbers and determine the sign of f ‘(x) for each of the intervals.

Step 4: Use the first derivative test to find the local maximum and minimum values.

f ‘(x) goes from negative to positive at x = –1, the First Derivative Test tells us that there is a local minimum at x = –1.
f (–1) = 2 is the local minimum value.

f ‘(x) goes from positive to negative at x = 0, the First Derivative Test tells us that there is a local maximum at x = 0.
f (0) = 3 is the local maximum value.

f ‘(x) goes from negative to positive at x = 1, the First Derivative Test tells us that there is a local minimum at x = 1.
f (1) = 2 is the local minimum value.

The Second Derivative Test

We can also use the Second Derivative Test to determine maximum or minimum values.

The Second Derivative Test

Suppose f ‘’ is continuous near c,

1. If f ‘(c) = 0 and f ‘’(c) > 0, then f has a local minimum at c.
2. If f ‘(c) = 0 and f ‘’(c) < 0, then f has a local maximum at c.

Example:

Use the Second Derivative Test to find the local maximum and minimum values of the function f(x) = x⁴ – 2x² + 3

Solution:

Step 1: Find the derivative of f

f ‘(x) = 4x³ – 4x = 4x(x² –1) = 4x(x –1)(x +1)

Step 2: Set f ‘(x) = 0 to get the critical numbers

f ‘(x) = 4x(x –1)(x +1) = 0
x = –1, 0, 1

Step 3: Find the second derivative

f ‘’(x) = 12x² – 4

Step 4: Evaluate f ‘’at the critical numbers

f ‘’(–1) = 8 > 0, so f (–1) = 2 is the local minimum value.
f ‘’(0) = – 4 < 0, so f (0) = 2 is the local maximum value.
f ‘’(1) = 8 > 0, so f (1) = 2 is the local minimum value.

Videos

Maxima and minima -
In this video, Professor Edward Burger examines maximum and minimum of functions. Find Maxima and Minima, Extreme Value Theorem, The Closed Interval Method

The second derivative test -
In this video, Professor Edward Burger discusses the second derivative and extreme points.

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