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

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

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

extreme points

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

Maximum (- 2, - 4), Minimum (2, 4)

Solution steps

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

Maximum (- 2, - 4)

Plug

x = 2

into

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

Minimum (2, 4)

Maximum (- 2, - 4), Minimum (2, 4)

f (x, y) = x^{3} - 6 x y + 3 y^{2} + 1

f (x) = \frac{x ^{2}}{x - 6}

f (x, y) = 4 x^{3} - 11 x y^{2} + 15 y + 12

f (x) = sin (x) + cos^{2} (x)

f (x) = cos (5 x)

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