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

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

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

extreme points

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

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

Solution steps

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

Maximum (- 2, - 2)

Plug

x = 2

into

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

Minimum (2, 2)

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

f (x) = (5 - x) e^{x}

f (x, y) = y^{3} - y x^{2} - 3 y^{2} \cdot x^{2} + 3 x^{4}

2 x^{2} - 9 x - 1

f (x) = x + \frac{16}{x}, 4 \leq x \leq 25

f (x) = 60 x^{3} - 120 x

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