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

The extreme f(x)=(x^2+x+1)/x is Maximum(-1,-1),Minimum(1,3)

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

extreme points

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

Maximum (- 1, - 1), Minimum (1, 3)

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} + x + 1}{x} : - 1

Maximum (- 1, - 1)

Plug

x = 1

into

\frac{x ^{2} + x + 1}{x} : 3

Minimum (1, 3)

Maximum (- 1, - 1), Minimum (1, 3)

(- x^{2} + 16) (x + 3)

f (x) = \frac{3 x ^{2}}{x ^{2} - 9}

y = \frac{3 x - 1}{2 x + 8}

f (x) = - 5

f (x) = \frac{1}{x + 4 - 2}

What is the extreme f(x)=(x^2+x+1)/x ?
The extreme f(x)=(x^2+x+1)/x is Maximum(-1,-1),Minimum(1,3)