What is the extreme f(x)=2x^4-4x^2+6 ?

The extreme f(x)=2x^4-4x^2+6 is Minimum(-1,4),Maximum(0,6),Minimum(1,4)

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extreme f(x)=2x^4-4x^2+6

extreme points

f (x) = 2 x^{4} - 4 x^{2} + 6

Minimum (- 1, 4), Maximum (0, 6), Minimum (1, 4)

Solution steps

f^{'} (x) = 8 x^{3} - 8 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

2 x^{4} - 4 x^{2} + 6 : 4

Minimum (- 1, 4)

Plug

x = 0

into

2 x^{4} - 4 x^{2} + 6 : 6

Maximum (0, 6)

Plug

x = 1

into

2 x^{4} - 4 x^{2} + 6 : 4

Minimum (1, 4)

Minimum (- 1, 4), Maximum (0, 6), Minimum (1, 4)

f (x) = x e^{7 x^{2}}

f (x, y) = 5 x + 3 y + 2

f (x) = ln (5 - 3 x^{2})

f (x) = \frac{x ^{3} + 3}{2 x ^{2} - 6 x - 4}

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

What is the extreme f(x)=2x^4-4x^2+6 ?
The extreme f(x)=2x^4-4x^2+6 is Minimum(-1,4),Maximum(0,6),Minimum(1,4)