how to find the absolute extrema of the function

Question

Find the absolute extrema of the function on the closed interval. (Order your answers from smallest to largest x, then from smallest to largest y.)

f(x) = sin(4x),    [0, π]

Answer 1

Candidates for absolute extrema include critical points and endpoints. Critical points occur when f'(x) = 0 and endpoints are given.

Endpoints: 0 and π

Critical Points:

Solve for f'(x) = 0

f'(x) = 4cos(4x)

0 = 4cos(4x)

cos(4x) = 0



Critical points occur at x = π/8, 3π/8, 5π/8, 7π/8

Now we need to test every point in our original function to see which is the greatest and smallest to determine our extrema.

(0, 0)

(π/8, 1)

(3π/8, -1)

(5π/8, 1)

(7π/8, -1)

(π, 0)

Out of these values, 1 is the largest and -1 is the smallest. Thus, our absolute extrema are:

Maxima: (π/8, 1) and (5π/8, 1)

Minima: (3π/8, -1) and (7π/8, -1)

Answer 2

The absolute extrema can only occur at an endpoint or critical point that lies within the interval.

f'(x) = 4cos(4x). f'(x) = 0 when 4x = π/2 + kπ

x = π/8 + kπ/4, k = 0, ±1, ±2, ...

When k = 0, we get x = π/8

When k = 1, we get x = 3π/8

When k = 2, we get x = 5π/8

When k = 3, we get x = 7π/8

Other choices for k give values of x that lie outside of the interval [0, π].

Check endpoints and critical points:

f(0) = sin0 = 0

f(π) = sin(4π) = 0

f(π/8) = sin(π/2) = 1

f(3π/8) = sin(3π/2) = -1

f(5π/8) = sin(5π/2) = 1

f(7π/8) = sin(7π/2) = -1

Absolute max is 1 and occurs when x = π/8 and 5π/8

Absolute min is -1 which occurs when x = 3π/8 and 7π/8