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

The extreme f(x)=((x^2-2x+1))/(x+1) is Maximum(-3,-8),Minimum(1,0)

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

extreme points

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

Maximum (- 3, - 8), Minimum (1, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 3

into

\frac{x ^{2} - 2 x + 1}{x + 1} : - 8

Maximum (- 3, - 8)

Plug

x = 1

into

\frac{x ^{2} - 2 x + 1}{x + 1} : 0

Minimum (1, 0)

Maximum (- 3, - 8), Minimum (1, 0)

f (x) = - x^{3} + 6 x^{2} - 9 x - 1

f (x) = (- \frac{1}{100} (x) (x - 300))

f (x) = 2 x + 6, - 4 \leq x \leq 2

f (x) = (x - 9)^{3}

f (x) = 1 - x^{\frac{4}{5}}

What is the extreme f(x)=((x^2-2x+1))/(x+1) ?
The extreme f(x)=((x^2-2x+1))/(x+1) is Maximum(-3,-8),Minimum(1,0)