What is the extreme f(x)=8x+2x^{-1} ?

The extreme f(x)=8x+2x^{-1} is Maximum(-1/2 ,-8),Minimum(1/2 ,8)

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extreme f(x)=8x+2x^{-1}

extreme points

f (x) = 8 x + 2 x^{- 1}

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

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Decreasing

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

Increasing

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

Plug

x = - \frac{1}{2}

into

8 x + 2 x^{- 1} : - 8

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

Plug

x = \frac{1}{2}

into

8 x + 2 x^{- 1} : 8

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

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

f (x) = 5 + 6 x^{2} - 4 x^{3}

f (x) = - 5 x^{2} + 8 x - 7

14 x (x^{2} - 4)^{6}

x^{4} - \frac{4}{3} x^{3} - 12 x^{2}

x^{2} - 4 x + 12

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