What is the extreme f(x)=x^{1/3}(x-7)^2 ?

The extreme f(x)=x^{1/3}(x-7)^2 is Maximum(1,36),Minimum(7,0)

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extreme f(x)=x^{1/3}(x-7)^2

extreme

Maximum (1, 36), Minimum (7, 0)

Solution steps

f^{'} (x) = \frac{7 ( x - 1 ) ( - 7 + x )}{3 x ^{\frac{2}{3}}}

Find intervals:

Increasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 1

into

x^{\frac{1}{3}} (x - 7)^{2} : 36

Maximum (1, 36)

Plug

x = 7

into

x^{\frac{1}{3}} (x - 7)^{2} : 0

Minimum (7, 0)

Maximum (1, 36), Minimum (7, 0)

What is the extreme f(x)=x^{1/3}(x-7)^2 ?
The extreme f(x)=x^{1/3}(x-7)^2 is Maximum(1,36),Minimum(7,0)