What is the extreme f(x)=(1+x)^2(6-x) ?

The extreme f(x)=(1+x)^2(6-x) is Minimum(-1,0),Maximum(11/3 , 1372/27)

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extreme f(x)=(1+x)^2(6-x)

extreme

Minimum (- 1, 0), Maximum (\frac{11}{3}, \frac{1372}{27})

Solution steps

f^{'} (x) = - (x + 1) (3 x - 11)

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < \frac{11}{3},

Decreasing

: \frac{11}{3} < x < \infty

Plug

x = - 1

into

(1 + x)^{2} (6 - x) : 0

Minimum (- 1, 0)

Plug

x = \frac{11}{3}

into

(1 + x)^{2} (6 - x) : \frac{1372}{27}

Maximum (\frac{11}{3}, \frac{1372}{27})

Minimum (- 1, 0), Maximum (\frac{11}{3}, \frac{1372}{27})

What is the extreme f(x)=(1+x)^2(6-x) ?
The extreme f(x)=(1+x)^2(6-x) is Minimum(-1,0),Maximum(11/3 , 1372/27)