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

The extreme f(x)=(2+x)^2(1-x)^2 is Minimum(-2,0),Maximum(-1/2 , 81/16),Minimum(1,0)

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

extreme points

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

Minimum (- 2, 0), Maximum (- \frac{1}{2}, \frac{81}{16}), Minimum (1, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - \frac{1}{2},

Decreasing

: - \frac{1}{2} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, 0)

Plug

x = - \frac{1}{2}

into

(2 + x)^{2} (1 - x)^{2} : \frac{81}{16}

Maximum (- \frac{1}{2}, \frac{81}{16})

Plug

x = 1

into

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

Minimum (1, 0)

Minimum (- 2, 0), Maximum (- \frac{1}{2}, \frac{81}{16}), Minimum (1, 0)

y = 6 x - ln (6 x)

f (x) = 7 + 81 x - 3 x^{3}

f (x, y) = 7 x - 8 y + 2 x y - x^{2} + y^{1}

f (x) = \frac{6}{x}

f (x, y) = (x - y) (9 - x y)

What is the extreme f(x)=(2+x)^2(1-x)^2 ?
The extreme f(x)=(2+x)^2(1-x)^2 is Minimum(-2,0),Maximum(-1/2 , 81/16),Minimum(1,0)