Thread: 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

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)

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.

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.

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

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

-Dan

try again ...

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