What is the extreme f(x)=2x^4-x^2+5 ?

The extreme f(x)=2x^4-x^2+5 is Minimum(-1/2 , 39/8),Maximum(0,5),Minimum(1/2 , 39/8)

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extreme f(x)=2x^4-x^2+5

extreme points

f (x) = 2 x^{4} - x^{2} + 5

Minimum (- \frac{1}{2}, \frac{39}{8}), Maximum (0, 5), Minimum (\frac{1}{2}, \frac{39}{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} + 5 : \frac{39}{8}

Minimum (- \frac{1}{2}, \frac{39}{8})

Plug

x = 0

into

2 x^{4} - x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = \frac{1}{2}

into

2 x^{4} - x^{2} + 5 : \frac{39}{8}

Minimum (\frac{1}{2}, \frac{39}{8})

Minimum (- \frac{1}{2}, \frac{39}{8}), Maximum (0, 5), Minimum (\frac{1}{2}, \frac{39}{8})

f (x) = 2 x^{2} + x + 2

y = 6 x^{4} + 8 x^{3}

πj z + 3 w

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

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

What is the extreme f(x)=2x^4-x^2+5 ?
The extreme f(x)=2x^4-x^2+5 is Minimum(-1/2 , 39/8),Maximum(0,5),Minimum(1/2 , 39/8)