What is the extreme f(x)=(x^2+4)(16-x^2) ?

The extreme f(x)=(x^2+4)(16-x^2) is Maximum(-sqrt(6),100),Minimum(0,64),Maximum(sqrt(6),100)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=(x^2+4)(16-x^2)

extreme

Maximum (- 6, 100), Minimum (0, 64), Maximum (6, 100)

Solution steps

f^{'} (x) = - 4 x (x + 6) (x - 6)

Find intervals:

Increasing

: - \infty < x < - 6,

Decreasing

: - 6 < x < 0,

Increasing

: 0 < x < 6,

Decreasing

: 6 < x < \infty

Plug

x = - 6

into

(x^{2} + 4) (16 - x^{2}) : 100

Maximum (- 6, 100)

Plug

x = 0

into

(x^{2} + 4) (16 - x^{2}) : 64

Minimum (0, 64)

Plug

x = 6

into

(x^{2} + 4) (16 - x^{2}) : 100

Maximum (6, 100)

Maximum (- 6, 100), Minimum (0, 64), Maximum (6, 100)

What is the extreme f(x)=(x^2+4)(16-x^2) ?
The extreme f(x)=(x^2+4)(16-x^2) is Maximum(-sqrt(6),100),Minimum(0,64),Maximum(sqrt(6),100)