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

The extreme f(x)=8x^2-x^4 is Maximum(-2,16),Minimum(0,0),Maximum(2,16)

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

extreme points

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

Maximum (- 2, 16), Minimum (0, 0), Maximum (2, 16)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = - 2

into

8 x^{2} - x^{4} : 16

Maximum (- 2, 16)

Plug

x = 0

into

8 x^{2} - x^{4} : 0

Minimum (0, 0)

Plug

x = 2

into

8 x^{2} - x^{4} : 16

Maximum (2, 16)

Maximum (- 2, 16), Minimum (0, 0), Maximum (2, 16)

f (t) = r (t) e^{- a t}

f (x) = 3 x^{4} + 4 x^{3} - 120 x^{2} + 120

f (x, y) = x^{2} + y^{2} + x y - 3 x - 6 y + 1

f (x) = x - x y + 2 y

What is the extreme f(x)=8x^2-x^4 ?
The extreme f(x)=8x^2-x^4 is Maximum(-2,16),Minimum(0,0),Maximum(2,16)