What is the extreme 2x^4-x^2+5 ?

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

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extreme 2x^4-x^2+5

extreme points

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) = 7 x^{9} - 9 x^{7} - 5 (- 3.4)

f (x) = - 3 + x^{2} - 3 x - 3 x^{3} + 2 x^{5} + 4 x^{4}

f (x) = - x^{3} + 6 x^{2} + 135 x + 1

x^{2} - 8 x + 15

x^{2} - 8 x + 13

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