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

The extreme f(x)=8x^2-4x^4 is Maximum(-1,4),Minimum(0,0),Maximum(1,4)

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

extreme points

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

Maximum (- 1, 4), Minimum (0, 0), Maximum (1, 4)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, 4)

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 1

into

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

Maximum (1, 4)

Maximum (- 1, 4), Minimum (0, 0), Maximum (1, 4)

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

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

f (x, y) = 2 x^{2} + y^{2} + 6 x y + 10 x - 6 y + 5

y = 2^{x} \cdot I n (5)

f (x) = 3 x^{4} - 4 x^{3} + 5

What is the extreme f(x)=8x^2-4x^4 ?
The extreme f(x)=8x^2-4x^4 is Maximum(-1,4),Minimum(0,0),Maximum(1,4)