What is the extreme f(x)=(8-x)(x+1)^2 ?

The extreme f(x)=(8-x)(x+1)^2 is Minimum(-1,0),Maximum(5,108)

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extreme f(x)=(8-x)(x+1)^2

extreme points

f (x) = (8 - x) (x + 1)^{2}

Minimum (- 1, 0), Maximum (5, 108)

Solution steps

f^{'} (x) = - 3 (x + 1) (x - 5)

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = - 1

into

(8 - x) (x + 1)^{2} : 0

Minimum (- 1, 0)

Plug

x = 5

into

(8 - x) (x + 1)^{2} : 108

Maximum (5, 108)

Minimum (- 1, 0), Maximum (5, 108)

f (x) = 5 x + 1 + \frac{1}{x - 1}

f (x) = x y^{2} - 4 x^{2} - 3 y^{2} + 12 x

4 x^{4} + 4 y^{4} - x y

f (x, y) = ln (x^{3} y + 3 x y^{3})

f (x) = 2 x ln (9 x)

What is the extreme f(x)=(8-x)(x+1)^2 ?
The extreme f(x)=(8-x)(x+1)^2 is Minimum(-1,0),Maximum(5,108)