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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 3)

Plug

x = 0

into

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

Minimum (0, 2)

Plug

x = 1

into

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

Maximum (1, 3)

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

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

f (x, y) = e^{- x - y}

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

f (x) = - x^{3} + 3 x^{2} + 45 x - 42

f (x) = (x^{2} + 12) (1 - x^{2})

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