What is the extreme f(x)=2x^4-x^2+1 ?

The extreme f(x)=2x^4-x^2+1 is Minimum(-1/2 , 7/8),Maximum(0,1),Minimum(1/2 , 7/8)

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extreme f(x)=2x^4-x^2+1

extreme points

f (x) = 2 x^{4} - x^{2} + 1

Minimum (- \frac{1}{2}, \frac{7}{8}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{7}{8})

Solution steps

f^{'} (x) = 8 x^{3} - 2 x

Find intervals:

Decreasing

: - \infty < x < - \frac{1}{2},

Increasing

: - \frac{1}{2} < x < 0,

Decreasing

: 0 < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{1}{2}

into

2 x^{4} - x^{2} + 1 : \frac{7}{8}

Minimum (- \frac{1}{2}, \frac{7}{8})

Plug

x = 0

into

2 x^{4} - x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = \frac{1}{2}

into

2 x^{4} - x^{2} + 1 : \frac{7}{8}

Minimum (\frac{1}{2}, \frac{7}{8})

Minimum (- \frac{1}{2}, \frac{7}{8}), Maximum (0, 1), Minimum (\frac{1}{2}, \frac{7}{8})

3 x^{3} + 2 y^{3} - 18 y^{2} - 9 x + 9

x^{4} - \frac{1}{2} x^{2} - 2

3 x^{5} - 5 x^{3} + 2

y = e^{- 4 t} + e^{- 5 t}

f (x) = \frac{2 x}{x ^{4} + 1}

What is the extreme f(x)=2x^4-x^2+1 ?
The extreme f(x)=2x^4-x^2+1 is Minimum(-1/2 , 7/8),Maximum(0,1),Minimum(1/2 , 7/8)