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

The extreme f(x)=(x^2+x+2)/(x-1) is Maximum(-1,-1),Minimum(3,7)

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

extreme points

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

Maximum (- 1, - 1), Minimum (3, 7)

Solution steps

f^{'} (x) = \frac{x ^{2} - 2 x - 3}{( x - 1 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 1

into

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

Maximum (- 1, - 1)

Plug

x = 3

into

\frac{x ^{2} + x + 2}{x - 1} : 7

Minimum (3, 7)

Maximum (- 1, - 1), Minimum (3, 7)

f (x) = - x^{2} - 4 x - 3

f (x) = \frac{12 x + 48}{7 x - 35}

y = \frac{x}{x + 2}

(0, 0), (2, - 3)

lo g_{e} (\frac{2 - x}{x + 3})

What is the extreme f(x)=(x^2+x+2)/(x-1) ?
The extreme f(x)=(x^2+x+2)/(x-1) is Maximum(-1,-1),Minimum(3,7)