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

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

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

extreme points

f (x) = 4 x^{\frac{1}{3}} - x \frac{4}{3}

Minimum (- 1, - \frac{8}{3}), Maximum (1, \frac{8}{3})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

4 x^{\frac{1}{3}} - x \frac{4}{3} : - \frac{8}{3}

Minimum (- 1, - \frac{8}{3})

Plug

x = 1

into

4 x^{\frac{1}{3}} - x \frac{4}{3} : \frac{8}{3}

Maximum (1, \frac{8}{3})

Minimum (- 1, - \frac{8}{3}), Maximum (1, \frac{8}{3})

f (x) = e^{x - 1}

\frac{x - 1}{x + 4}

(0, 0), (3, 0)

f (x) = x + 3 - 4

\frac{x ^{2} - 2 x + 4}{x - 2}

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