What is the extreme f(x)=15x^4-15x^2 ?

The extreme f(x)=15x^4-15x^2 is Minimum(-1/(sqrt(2)),-15/4),Maximum(0,0),Minimum(1/(sqrt(2)),-15/4)

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extreme f(x)=15x^4-15x^2

extreme points

f (x) = 15 x^{4} - 15 x^{2}

Minimum (- \frac{1}{2}, - \frac{15}{4}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{15}{4})

Solution steps

f^{'} (x) = 60 x^{3} - 30 x

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 0,

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

15 x^{4} - 15 x^{2} : - \frac{15}{4}

Minimum (- \frac{1}{2}, - \frac{15}{4})

Plug

x = 0

into

15 x^{4} - 15 x^{2} : 0

Maximum (0, 0)

Plug

x = \frac{1}{2}

into

15 x^{4} - 15 x^{2} : - \frac{15}{4}

Minimum (\frac{1}{2}, - \frac{15}{4})

Minimum (- \frac{1}{2}, - \frac{15}{4}), Maximum (0, 0), Minimum (\frac{1}{2}, - \frac{15}{4})

f (x) = \frac{2 x + 5}{x - 7}

f (x) = \frac{2 x - 8}{3 x ^{2} - 7 x - 20}

f (x) = \frac{3 x + 4}{x ^{2} + x - 6}

f (x) = \frac{1}{2 6 - x}

\frac{5 x}{x - 4}

What is the extreme f(x)=15x^4-15x^2 ?
The extreme f(x)=15x^4-15x^2 is Minimum(-1/(sqrt(2)),-15/4),Maximum(0,0),Minimum(1/(sqrt(2)),-15/4)