What is the extreme y=x^2(x^2-5) ?

The extreme y=x^2(x^2-5) is Minimum(-sqrt(5/2),-25/4),Maximum(0,0),Minimum(sqrt(5/2),-25/4)

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extreme y=x^2(x^2-5)

extreme points

y = x^{2} (x^{2} - 5)

Minimum (- \frac{5}{2}, - \frac{25}{4}), Maximum (0, 0), Minimum (\frac{5}{2}, - \frac{25}{4})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{5}{2},

Increasing

: - \frac{5}{2} < x < 0,

Decreasing

: 0 < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < \infty

Plug

x = - \frac{5}{2}

into

x^{2} (x^{2} - 5) : - \frac{25}{4}

Minimum (- \frac{5}{2}, - \frac{25}{4})

Plug

x = 0

into

x^{2} (x^{2} - 5) : 0

Maximum (0, 0)

Plug

x = \frac{5}{2}

into

x^{2} (x^{2} - 5) : - \frac{25}{4}

Minimum (\frac{5}{2}, - \frac{25}{4})

Minimum (- \frac{5}{2}, - \frac{25}{4}), Maximum (0, 0), Minimum (\frac{5}{2}, - \frac{25}{4})

f (x) = 6 x + 18

f (x) = \frac{4 y + 7}{3 - x}

3 cos^{2} (x), 0 \leq x \leq π

f (x) = x + \frac{9}{x} + 2, 1 \leq x \leq 18

f (x) = 7 x^{4} - 6 x^{2} - 1

What is the extreme y=x^2(x^2-5) ?
The extreme y=x^2(x^2-5) is Minimum(-sqrt(5/2),-25/4),Maximum(0,0),Minimum(sqrt(5/2),-25/4)