What is the extreme f(x)=-3x^4+18x^2-15 ?

The extreme f(x)=-3x^4+18x^2-15 is Maximum(-sqrt(3),12),Minimum(0,-15),Maximum(sqrt(3),12)

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extreme f(x)=-3x^4+18x^2-15

extreme points

f (x) = - 3 x^{4} + 18 x^{2} - 15

Maximum (- 3, 12), Minimum (0, - 15), Maximum (3, 12)

Solution steps

f^{'} (x) = - 12 x^{3} + 36 x

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = - 3

into

- 3 x^{4} + 18 x^{2} - 15 : 12

Maximum (- 3, 12)

Plug

x = 0

into

- 3 x^{4} + 18 x^{2} - 15 : - 15

Minimum (0, - 15)

Plug

x = 3

into

- 3 x^{4} + 18 x^{2} - 15 : 12

Maximum (3, 12)

Maximum (- 3, 12), Minimum (0, - 15), Maximum (3, 12)

- x + 4

a x^{2} - 4 x + 2 a

f (x) = (x^{2} - 1) (x^{2} - 4) (x^{2} - 9)

f (x) = \frac{x + 5}{x - 3}

\frac{3}{x ^{2} - 4}

What is the extreme f(x)=-3x^4+18x^2-15 ?
The extreme f(x)=-3x^4+18x^2-15 is Maximum(-sqrt(3),12),Minimum(0,-15),Maximum(sqrt(3),12)