What is the extreme f(x)=(x^2+4)/(8x) ?

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

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extreme f(x)=(x^2+4)/(8x)

extreme points

f (x) = \frac{x ^{2} + 4}{8 x}

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

Solution steps

f^{'} (x) = \frac{x ^{2} - 4}{8 x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

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

Plug

x = 2

into

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

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

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

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