What is the extreme f(x)=4x-12x^{1/3} ?

The extreme f(x)=4x-12x^{1/3} is Minimum(-1,8),Maximum(1,-8)

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extreme f(x)=4x-12x^{1/3}

extreme points

f (x) = 4 x - 12 x^{\frac{1}{3}}

Minimum (- 1, 8), Maximum (1, - 8)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

4 x - 12 x^{\frac{1}{3}} : 8

Minimum (- 1, 8)

Plug

x = 1

into

4 x - 12 x^{\frac{1}{3}} : - 8

Maximum (1, - 8)

Minimum (- 1, 8), Maximum (1, - 8)

f (x) = 3 x + 2 cos (x)

f (t) = \frac{t}{t - 4}

f (x) = - 2 x^{\frac{5}{3}} + 4 3 x

f (x) = \frac{( x ^{2} - 5 x + 4 )}{x - 5}

f (x) = 2 x^{3} - 3 x^{2} - 72 x + 15

What is the extreme f(x)=4x-12x^{1/3} ?
The extreme f(x)=4x-12x^{1/3} is Minimum(-1,8),Maximum(1,-8)