Curve Sketching Tangents, etc Question

• May 31st 2007, 06:52 PM
SportfreundeKeaneKent
Curve Sketching Tangents, etc Question
For the first question, I get x=1, -1 for critical points. The interval of increase is -1<x<1. The local extrema are (-1,-1/2) and (1,1/2)
The H.A is y=0 and the curve crosses it at (0,0)? I could be wrong, here's the question differentiated-f'(x)=1-x^2/(1-x^2)^2

For the second one, I get the slope to equal 1/-5 and y intercept to be -14/5?

http://img518.imageshack.us/img518/1817/23rf2.png
• May 31st 2007, 07:07 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
For the first question, I get x=1, -1 for critical points. The interval of increase is -1<x<1. The local extrema are (-1,-1/2) and (1,1/2)
The H.A is y=0 and the curve crosses it at (0,0)? I could be wrong, here's the question differentiated-f'(x)=1-x^2/(1-x^2)^2

For the second one, I get the slope to equal 1/-5 and y intercept to be -14/5?

http://img518.imageshack.us/img518/1817/23rf2.png

yup. everything is correct except the derivative formula you have for the first one, but i'm sure that was just a typo and you actually did it correctly...just to make sure, you mind telling me the correct function for f'(x)?
• Jun 1st 2007, 06:16 AM
curvature
curve sketching
The curve is shown as follows:
• Jun 1st 2007, 06:39 AM
curvature
Tangent to the ellipse
Tangent to the ellipse
• Jun 1st 2007, 02:37 PM
SportfreundeKeaneKent
It goes from f(x)=x/1+x^2 to being differentiated to

f'(x)=(1-x^2)-x(2x)/(1+x^2)^2
• Jun 1st 2007, 02:40 PM
Jhevon
Quote:

Originally Posted by SportfreundeKeaneKent
It goes from f(x)=x/1+x^2 to being differentiated to

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

that is incorrect, and you should put brackets around the numerator

it would be:

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