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
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)