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

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

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

extreme points

\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)

f (x) = 2 x^{2} - 6 x + 4, 0 \leq x \leq 10

x^{2} + y^{2} + x y

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

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

12 y^{3} + 64 x^{2} y - 2304 y - 7

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