What is the extreme f(x)=12+2x^2-x^4 ?

The extreme f(x)=12+2x^2-x^4 is Maximum(-1,13),Minimum(0,12),Maximum(1,13)

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extreme f(x)=12+2x^2-x^4

extreme points

f (x) = 12 + 2 x^{2} - x^{4}

Maximum (- 1, 13), Minimum (0, 12), Maximum (1, 13)

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

12 + 2 x^{2} - x^{4} : 13

Maximum (- 1, 13)

Plug

x = 0

into

12 + 2 x^{2} - x^{4} : 12

Minimum (0, 12)

Plug

x = 1

into

12 + 2 x^{2} - x^{4} : 13

Maximum (1, 13)

Maximum (- 1, 13), Minimum (0, 12), Maximum (1, 13)

f (x) = - 15 + 7 x - x^{2}

f (x) = \frac{0.002 x ^{2} + 24}{x}

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

f (x) = - 2 + x - 5 x^{2}

f (x) = 4 sin (x) + 4 cos (x), 0 \leq x \leq 2 π

What is the extreme f(x)=12+2x^2-x^4 ?
The extreme f(x)=12+2x^2-x^4 is Maximum(-1,13),Minimum(0,12),Maximum(1,13)