What is the extreme f(x)=((x^2-5)^3)/(125) ?

The extreme f(x)=((x^2-5)^3)/(125) is Minimum(0.00000…,-1.00000…)

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extreme f(x)=((x^2-5)^3)/(125)

extreme points

f (x) = \frac{( x ^{2} - 5 ) ^{3}}{125}

Minimum (0.00000 \dots, - 1.00000 \dots)

Solution steps

f^{'} (x) = \frac{6 x ( x ^{2} - 5 ) ^{2}}{125}

Find intervals:

Decreasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 0.00000 \dots

into

\frac{( x ^{2} - 5 ) ^{3}}{125} : - 1.00000 \dots

Minimum (0.00000 \dots, - 1.00000 \dots)

4 cos (x)

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 5

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

f (x) = - 2 x + 5 ln (2 x)

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

What is the extreme f(x)=((x^2-5)^3)/(125) ?
The extreme f(x)=((x^2-5)^3)/(125) is Minimum(0.00000…,-1.00000…)