What is the extreme (x^2+3x)/(x-1) ?

The extreme (x^2+3x)/(x-1) is Maximum(-1,1),Minimum(3,9)

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extreme (x^2+3x)/(x-1)

extreme points

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

Maximum (- 1, 1), Minimum (3, 9)

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} + 3 x}{x - 1} : 1

Maximum (- 1, 1)

Plug

x = 3

into

\frac{x ^{2} + 3 x}{x - 1} : 9

Minimum (3, 9)

Maximum (- 1, 1), Minimum (3, 9)

f (x) = \frac{x}{x ^{2} + 12 x + 32}

f (x) = 18 x + 3 x^{2} - 4 x^{3}

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

\frac{1}{2} sin (x + \frac{π}{4})

f (x) = x^{\frac{6}{7}} - 7

What is the extreme (x^2+3x)/(x-1) ?
The extreme (x^2+3x)/(x-1) is Maximum(-1,1),Minimum(3,9)