# Thread: Finding tangent line of a function

1. ## Finding tangent line of a function

Hi there -- i could use a hand with this function:

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

I'm asked to find the tangent line at x=3, and find the horizontal tangents, (where, if I remember, the tangent line = zero, correct?). I've been looking at this for a while, can't seem to move it along.

thanks,
Renee

2. Originally Posted by rhunt
Hi there -- i could use a hand with this function:

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

I'm asked to find the tangent line at x=3, and find the horizontal tangents, (where, if I remember, the tangent line = zero, correct?). I've been looking at this for a while, can't seem to move it along.

thanks,
Renee
$\displaystyle f(x)=\frac{2x}{1+x^2}$

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

3. Originally Posted by rhunt
Hi there -- i could use a hand with this function:

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

I'm asked to find the tangent line at x=3, and find the horizontal tangents, (where, if I remember, the tangent line = zero, correct?).
No, not "tangent line= zero". A line is not equal to a number. The slope of a horizontal line is 0.

I've been looking at this for a while, can't seem to move it along.

thanks,
Renee
Now that alexmahone has given you the derivative of the function, can you do it?

4. I think so. Solving for the horizontal slope = 0...you do by setting fprime(x) equal to zero and then solving for x, correct?

5. If f(x)=2x/(1+x^2) then

f'(x)=2(1+x^2)-(4x^2)/(x^2+1)^2
f'(x)=-2x^2+2/(x^2+1)^2

Then, you will f'(x) of 3:

f'(3)=-2(3)^2+2/(3^2+1)^2
f'(3)=36+2/100
f'(3)=19/50

This is now the slope of the line at x=3. You can then use point-slope form to find the equation of the line tangent, but first you need to find the y coordinate using the function f(x) and finding f(3) which results in 3/5.

Then, the equation of the line tangent to will be:

y-3/5=(19/50)(x-3)