What is the extreme f(x)=4x-48x^{1/3} ?

The extreme f(x)=4x-48x^{1/3} is Minimum(-8,64),Maximum(0,0),Minimum(8,-64)

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extreme f(x)=4x-48x^{1/3}

extreme

Minimum (- 8, 64), Maximum (0, 0), Minimum (8, - 64)

Solution steps

f^{'} (x) = 4 - \frac{16}{x ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{8},

Increasing

: - \frac{1}{8} < x < \frac{1}{8},

Decreasing

: \frac{1}{8} < x < \infty

Plug

x = - 8

into

4 x - 48 x^{\frac{1}{3}} : 64

Minimum (- 8, 64)

Plug

x = 0

into

4 x - 48 x^{\frac{1}{3}} : 0

Maximum (0, 0)

Plug

x = 8

into

4 x - 48 x^{\frac{1}{3}} : - 64

Minimum (8, - 64)

Minimum (- 8, 64), Maximum (0, 0), Minimum (8, - 64)

What is the extreme f(x)=4x-48x^{1/3} ?
The extreme f(x)=4x-48x^{1/3} is Minimum(-8,64),Maximum(0,0),Minimum(8,-64)