What is the extreme f(x)=(x^2+1)/x ?

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

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extreme f(x)=(x^2+1)/x

extreme points

f (x) = \frac{x ^{2} + 1}{x}

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 2)

Plug

x = 1

into

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

Minimum (1, 2)

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

\frac{1}{x + 6}

(x - 2) (x + 4) (x - 6) (x + 8)

f (x) = \frac{x + 7}{2 - x}

f (x) = \frac{( x ^{2} - x )}{x ^{2} - 9 x + 8}

(- 6, 5), (- 3, - 3)

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