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

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

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

extreme points

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

Maximum (- 2, - 1), Minimum (2, 1)

Solution steps

f^{'} (x) = \frac{x ^{2} - 4}{4 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}{4 x} : - 1

Maximum (- 2, - 1)

Plug

x = 2

into

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

Minimum (2, 1)

Maximum (- 2, - 1), Minimum (2, 1)

d o main f (x) = - 6 x^{2}

d o main \frac{x + 3}{x - 9}

a sy m pt o t es y = \frac{x}{x ^{2} - 4}

d o main \frac{x ^{2}}{x ^{2} - 9}

sy mm e t ry x^{2} - 8 x

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