What is the extreme f(x)=x^4-8x^2+16[-1.4] ?

The extreme f(x)=x^4-8x^2+16[-1.4] is Minimum(-2,-38.4),Maximum(0,-22.4),Minimum(2,-38.4)

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extreme f(x)=x^4-8x^2+16[-1.4]

extreme points

f (x) = x^{4} - 8 x^{2} + 16 [- 1.4]

Minimum (- 2, - 38.4), Maximum (0, - 22.4), Minimum (2, - 38.4)

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} + 16 [- 1.4] : - 38.4

Minimum (- 2, - 38.4)

Plug

x = 0

into

x^{4} - 8 x^{2} + 16 [- 1.4] : - 22.4

Maximum (0, - 22.4)

Plug

x = 2

into

x^{4} - 8 x^{2} + 16 [- 1.4] : - 38.4

Minimum (2, - 38.4)

Minimum (- 2, - 38.4), Maximum (0, - 22.4), Minimum (2, - 38.4)

f (x) = 3 x - x^{3} + 5

x^{3} + 3 x^{2} - 16

f (x) = 2 cos (x) - sin (2 x)

+ \frac{6 x ^{2} + 35 x - 6}{4 x ^{2} + 23 x - 6}

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

What is the extreme f(x)=x^4-8x^2+16[-1.4] ?
The extreme f(x)=x^4-8x^2+16[-1.4] is Minimum(-2,-38.4),Maximum(0,-22.4),Minimum(2,-38.4)