What is the extreme f(x)=3x-9x^{1/3} ?

The extreme f(x)=3x-9x^{1/3} is Minimum(-1,6),Maximum(1,-6)

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extreme f(x)=3x-9x^{1/3}

extreme points

f (x) = 3 x - 9 x^{\frac{1}{3}}

Minimum (- 1, 6), Maximum (1, - 6)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

3 x - 9 x^{\frac{1}{3}} : 6

Minimum (- 1, 6)

Plug

x = 1

into

3 x - 9 x^{\frac{1}{3}} : - 6

Maximum (1, - 6)

Minimum (- 1, 6), Maximum (1, - 6)

g (x, y) = c x + y x

x^{3} + 6 x^{2} + 9 x

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

f (x) = 2 x^{3} - 3 x^{2} - 36 x + 1

f (x, y) = 2 x^{2} + 4 x y^{3} - 2 y^{2} + 15

What is the extreme f(x)=3x-9x^{1/3} ?
The extreme f(x)=3x-9x^{1/3} is Minimum(-1,6),Maximum(1,-6)