What is the extreme f(x)=(x^2+9)/(2x) ?

The extreme f(x)=(x^2+9)/(2x) is Maximum(-3,-3),Minimum(3,3)

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extreme f(x)=(x^2+9)/(2x)

extreme points

f (x) = \frac{x ^{2} + 9}{2 x}

Maximum (- 3, - 3), Minimum (3, 3)

Solution steps

f^{'} (x) = \frac{x ^{2} - 9}{2 x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{x ^{2} + 9}{2 x} : - 3

Maximum (- 3, - 3)

Plug

x = 3

into

\frac{x ^{2} + 9}{2 x} : 3

Minimum (3, 3)

Maximum (- 3, - 3), Minimum (3, 3)

f (x) = x - y = 10

w (x, y) = x y + \frac{e ^{y}}{y ^{2} + 1}

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

f (x, y) = 9 - x^{2} - 4 - y^{2}

f (x) = x x + 9

What is the extreme f(x)=(x^2+9)/(2x) ?
The extreme f(x)=(x^2+9)/(2x) is Maximum(-3,-3),Minimum(3,3)