# Finding open intervals of a function and extremas

• Nov 18th 2012, 11:24 AM
Chaim
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) = (x2 - 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) + (x2 - 2x + 1)(1))/(x + 1)2 which is simplified into (2x2 + 2x - 2x - 2 + x2 - 2x + 1)/(x + 1)2
Which is simplified into: (3x2 - 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

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)
• Nov 18th 2012, 12:40 PM
topsquark
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) = (x2 - 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.
$\lim_{x \to \infty}\frac{x^2 - 2x + 1}{x+1} = \lim_{x \to \infty} \left [ x - 3+ \frac{4}{x+1} \right ]$
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) + (x2 - 2x + 1)(1))/(x + 1)2 which is simplified into (2x2 + 2x - 2x - 2 + x2 - 2x + 1)/(x + 1)2
Which is simplified into: (3x2 - 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 $f'(x) = \frac{x^2 + 2x - 3}{(x + 1)^2} = \frac{(x - 1)(x + 3)}{(x + 1)^2}$

-Dan
• Nov 18th 2012, 12:41 PM
skeeter
Re: Finding open intervals of a function and extremas
$f(x) = \frac{x^2-2x+1}{x+1}$

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

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

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

try again ...
• Nov 18th 2012, 03:56 PM
Chaim
Re: Finding open intervals of a function and extremas
Ah thanks! That makes more sense now for the critical numbers and sign chart!