What is the extreme f(x)= 3/7 (x^2-25)^{2/3} ?

The extreme f(x)= 3/7 (x^2-25)^{2/3} is Minimum(-5,0),Maximum(0,(3(-25)^{2/3})/7),Minimum(5,0)

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extreme f(x)= 3/7 (x^2-25)^{2/3}

extreme points

f (x) = \frac{3}{7} (x^{2} - 25)^{\frac{2}{3}}

Minimum (- 5, 0), Maximum (0, \frac{3 ( - 25 ) ^{\frac{2}{3}}}{7}), Minimum (5, 0)

Solution steps

f^{'} (x) = \frac{4 x}{7 ( x ^{2} - 25 ) ^{\frac{1}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Increasing

: - 5 < x < 0,

Decreasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 5

into

\frac{3}{7} (x^{2} - 25)^{\frac{2}{3}} : 0

Minimum (- 5, 0)

Plug

x = 0

into

\frac{3}{7} (x^{2} - 25)^{\frac{2}{3}} : \frac{3 ( - 25 ) ^{\frac{2}{3}}}{7}

Maximum (0, \frac{3 ( - 25 ) ^{\frac{2}{3}}}{7})

Plug

x = 5

into

\frac{3}{7} (x^{2} - 25)^{\frac{2}{3}} : 0

Minimum (5, 0)

Minimum (- 5, 0), Maximum (0, \frac{3 ( - 25 ) ^{\frac{2}{3}}}{7}), Minimum (5, 0)

- 7 x^{2} + 14 x + 245

f (x) = 15 x + 20 y

f (x) = x^{3} - 6 x^{2} + 12 x + 15

f x 4 - x

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

What is the extreme f(x)= 3/7 (x^2-25)^{2/3} ?
The extreme f(x)= 3/7 (x^2-25)^{2/3} is Minimum(-5,0),Maximum(0,(3(-25)^{2/3})/7),Minimum(5,0)