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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, - 16)

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 2

into

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

Minimum (2, - 16)

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

f (x) = (x^{3} - 3 x)^{\frac{1}{2}}

f (t) = 108 t - t^{3}, - 7 \leq t < \infty

f (x) = 2 ln (x) 1 + 3 ln (x) 2 + 3 ln (x) 3

f (x) = ln (x), 0 < x \leq 4

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

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