Finding open intervals of a function and extremas

Find all open intervals on which the function is increasing or decreasing, and locate all relative extrema. Graph it.

**f(x) = (x**^{2 }- 2x + 1 )/(x + 1)

What I did:

I know the **veritcal asymptote is -1, which is the discontinuity**

I know the horizontal asymptote is 1 since the factor for the top is (x + 1)(x - 1)

The derivative is: ((x + 1)(2x - 2) + (x^{2} - 2x + 1)(1))/(x + 1)^{2 }which is simplified into (2x^{2 }+ 2x - 2x - 2 + x^{2} - 2x + 1)/(x + 1)^{2}

Which is simplified into: (3x^{2} - 2x - 1)/(x + 1)^{2}

The numerator is factored into (3x + 1)(x - 1), **but do I get my critical numbers from this numerator? **

Because it would be x=(-1/3) and 1, which is different from the answer

Also does anyone know how to do the 'sign chart?' That way I could get my minimum and maximum.

An EXAMPLE of a sign chart is:

x (-∞, -3), -3x, (-3, 0), 0, (0, 3), 3, (3, ∞),

f(x) positive, undefined, positive, 0, negative, undefined, negative

Answers are:

Critical numbers: x = -3, 1

Discontinuity: x = -1

Increasing on (-∞, -3) and (1, ∞)

Decreasing on (-3, -1) and (-1, 1)

Relative maximum (-3, -8)

Relative minimum: (1, 0)

Re: Finding open intervals of a function and extremas

Quote:

Originally Posted by

**Chaim** Find all open intervals on which the function is increasing or decreasing, and locate all relative extrema. Graph it.

**f(x) = (x**^{2 }- 2x + 1 )/(x + 1)

What I did:

I know the **veritcal asymptote is -1, which is the discontinuity**

Yes.

Quote:

Originally Posted by

**Chaim** I know the horizontal asymptote is 1 since the factor for the top is (x + 1)(x - 1)

Double check this.

$\displaystyle \lim_{x \to \infty}\frac{x^2 - 2x + 1}{x+1} = \lim_{x \to \infty} \left [ x - 3+ \frac{4}{x+1} \right ]$

which does not fall to zero at infinity. This is actually a *slant* asymptote.

Quote:

Originally Posted by

**Chaim** The derivative is: ((x + 1)(2x - 2) + (x^{2} - 2x + 1)(1))/(x + 1)^{2 }which is simplified into (2x^{2 }+ 2x - 2x - 2 + x^{2} - 2x + 1)/(x + 1)^{2}

Which is simplified into: (3x^{2} - 2x - 1)/(x + 1)^{2}

The numerator is factored into (3x + 1)(x - 1), [B]but do I get my critical numbers from this numerator?

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

-Dan

Re: Finding open intervals of a function and extremas

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

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

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

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

try again ...

Re: Finding open intervals of a function and extremas

Ah thanks! That makes more sense now for the critical numbers and sign chart!