# Thread: Curve Sketching Tangents, etc Question

1. ## 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?

2. 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?

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)?

3. ## curve sketching

The curve is shown as follows:

4. ## Tangent to the ellipse

Tangent to the ellipse

5. It goes from f(x)=x/1+x^2 to being differentiated to

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

6. 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:

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