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

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

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

extreme points

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

Maximum (- 2, 8), Minimum (0, 0), Maximum (2, 8)

Solution steps

f^{'} (x) = 16 x - 8 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} - 2 x^{4} : 8

Maximum (- 2, 8)

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 2

into

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

Maximum (2, 8)

Maximum (- 2, 8), Minimum (0, 0), Maximum (2, 8)

P (q, s) = q + 2 + s

f (x) = \frac{5}{3} x^{3} - \frac{65}{2} x^{2} + 200 x - 1

f (x) = 4 x^{2} e^{- 0.25 x}

f (x) = x^{3} - 9 x^{2} + 15 x + 2

f (x) = (x^{2} - 1) e^{- x}

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