What is the extreme f(x)= 6/7 (x^2-9)^{2/3} ?

The extreme f(x)= 6/7 (x^2-9)^{2/3} is Minimum(-3,0),Maximum(0,(6(-9)^{2/3})/7),Minimum(3,0)

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extreme f(x)= 6/7 (x^2-9)^{2/3}

extreme points

f (x) = \frac{6}{7} (x^{2} - 9)^{\frac{2}{3}}

Minimum (- 3, 0), Maximum (0, \frac{6 ( - 9 ) ^{\frac{2}{3}}}{7}), Minimum (3, 0)

Solution steps

f^{'} (x) = \frac{8 x}{7 ( x ^{2} - 9 ) ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{6}{7} (x^{2} - 9)^{\frac{2}{3}} : 0

Minimum (- 3, 0)

Plug

x = 0

into

\frac{6}{7} (x^{2} - 9)^{\frac{2}{3}} : \frac{6 ( - 9 ) ^{\frac{2}{3}}}{7}

Maximum (0, \frac{6 ( - 9 ) ^{\frac{2}{3}}}{7})

Plug

x = 3

into

\frac{6}{7} (x^{2} - 9)^{\frac{2}{3}} : 0

Minimum (3, 0)

Minimum (- 3, 0), Maximum (0, \frac{6 ( - 9 ) ^{\frac{2}{3}}}{7}), Minimum (3, 0)

f (x, y) = 7 x y + 14 x - x^{2} + 2 y^{2}

4 x^{2} - 8 x^{4}

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

f (x) = 9 x^{3} + 21 x^{2} + 8 x - 3

f (x) = x^{2} - 2 x + 5 [- 1.4]

What is the extreme f(x)= 6/7 (x^2-9)^{2/3} ?
The extreme f(x)= 6/7 (x^2-9)^{2/3} is Minimum(-3,0),Maximum(0,(6(-9)^{2/3})/7),Minimum(3,0)