What is the extreme f(x)=x^4-8x^2+9 ?

The extreme f(x)=x^4-8x^2+9 is Minimum(-2,-7),Maximum(0,9),Minimum(2,-7)

extreme f(x)=x^4-8x^2+9

extreme points

f (x) = x^{4} - 8 x^{2} + 9

Minimum (- 2, - 7), Maximum (0, 9), Minimum (2, - 7)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 8 x^{2} + 9 : - 7

Minimum (- 2, - 7)

Plug

x = 0

into

x^{4} - 8 x^{2} + 9 : 9

Maximum (0, 9)

Plug

x = 2

into

x^{4} - 8 x^{2} + 9 : - 7

Minimum (2, - 7)

Minimum (- 2, - 7), Maximum (0, 9), Minimum (2, - 7)

f (x) = ln (5 x^{2} - 4 x - 3)

f (x) = x^{- 2} - 13 x^{- 1}

f (x) = 68 x - x^{3}

f (x) = 2 x^{3} - 7

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

What is the extreme f(x)=x^4-8x^2+9 ?
The extreme f(x)=x^4-8x^2+9 is Minimum(-2,-7),Maximum(0,9),Minimum(2,-7)