What is the extreme f(x)=-x^4+4x^2-1 ?

The extreme f(x)=-x^4+4x^2-1 is Maximum(-sqrt(2),3),Minimum(0,-1),Maximum(sqrt(2),3)

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extreme f(x)=-x^4+4x^2-1

extreme points

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

Maximum (- 2, 3), Minimum (0, - 1), Maximum (2, 3)

Solution steps

f^{'} (x) = - 4 x^{3} + 8 x

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

- x^{4} + 4 x^{2} - 1 : 3

Maximum (- 2, 3)

Plug

x = 0

into

- x^{4} + 4 x^{2} - 1 : - 1

Minimum (0, - 1)

Plug

x = 2

into

- x^{4} + 4 x^{2} - 1 : 3

Maximum (2, 3)

Maximum (- 2, 3), Minimum (0, - 1), Maximum (2, 3)

f (x) = 9 + 54 x - 2 x^{3}

f (x) = 5^{x}

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

f (x) = \frac{x}{2} - 4

What is the extreme f(x)=-x^4+4x^2-1 ?
The extreme f(x)=-x^4+4x^2-1 is Maximum(-sqrt(2),3),Minimum(0,-1),Maximum(sqrt(2),3)