What is the extreme g(x)=-2x^4+8x^2-6 ?

The extreme g(x)=-2x^4+8x^2-6 is Maximum(-sqrt(2),2),Minimum(0,-6),Maximum(sqrt(2),2)

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extreme g(x)=-2x^4+8x^2-6

extreme points

g (x) = - 2 x^{4} + 8 x^{2} - 6

Maximum (- 2, 2), Minimum (0, - 6), Maximum (2, 2)

Solution steps

f^{'} (x) = - 8 x^{3} + 16 x

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

- 2 x^{4} + 8 x^{2} - 6 : 2

Maximum (- 2, 2)

Plug

x = 0

into

- 2 x^{4} + 8 x^{2} - 6 : - 6

Minimum (0, - 6)

Plug

x = 2

into

- 2 x^{4} + 8 x^{2} - 6 : 2

Maximum (2, 2)

Maximum (- 2, 2), Minimum (0, - 6), Maximum (2, 2)

\frac{4 x ^{3} - 3 x ^{2} + 2 + x}{x ( 2 x ^{2} - 3 x + 1 )}

\frac{( x - 6 )}{x + 6}

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

f (x) = x^{3} + 3 x^{2} - 16 x - 48

3 - 2 x - x^{3}

What is the extreme g(x)=-2x^4+8x^2-6 ?
The extreme g(x)=-2x^4+8x^2-6 is Maximum(-sqrt(2),2),Minimum(0,-6),Maximum(sqrt(2),2)