What is the extreme f(x)=(8-(\sqrt[3]{x^2+2x+1})) ?

The extreme f(x)=(8-(\sqrt[3]{x^2+2x+1})) is Maximum(-1,8)

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extreme f(x)=(8-(\sqrt[3]{x^2+2x+1}))

extreme

Maximum (- 1, 8)

Solution steps

f^{'} (x) = - \frac{2 ( x + 1 )}{3 ( ( x + 1 ) ^{2} ) ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \infty

Plug

x = - 1

into

8 - 3 x^{2} + 2 x + 1 : 8

Maximum (- 1, 8)

What is the extreme f(x)=(8-(\sqrt[3]{x^2+2x+1})) ?
The extreme f(x)=(8-(\sqrt[3]{x^2+2x+1})) is Maximum(-1,8)