Question

Find the equation of the line tangent to f(x)=x/(x^2+3) at point 0,0

Answer 1

So the equation of our tangent line will be of the form,\[\Large y=mx+b\]
Where \(\large m\) is the slope of the tangent line.\[\Large f'(0)=m\]

So it looks like we want to start by taking the derivative of the function.
Have you tried that yet?

Answer 2

yes
I got x^2+3-2x/x^4+6x^2+9

Answer 3

I used the quotient rule, but that's where I lost it. I didn't think it should be this messy

Answer 4

\[\large f(x)=\frac{x}{x^2+3}\]

\[\large f'(x)=\frac{\color{royalblue}{(x)'}(x^2+3)-(x)\color{royalblue}{(x^2+3)'}}{(x^2+3)^2}\]
\[\large f'(x)=\frac{\color{orangered}{(1)}(x^2+3)-(x)\color{orangered}{(2x)}}{(x^2+3)^2}\]Small mistake on the top, see how our term on the right should be 2x^2?
And also, DO NOT expand out the denominator! :O
Let's leave that alone for now.

Answer 5

ahh yes I see, ok so that brings us to 3-x^2/(x^2+3)^2

Answer 6

\[\large f'(x)=\frac{3-x^2}{(x^2+3)^2}\]
Ok good :)
So from here, we want the value of the derivative at x=0.

\[\Large f'(0)=?\]

Answer 7

x=sqrt3

Answer 8

wait sorry I mean 3/9

Answer 9

or 1/3

Answer 10

\[\large f'(0)=\frac{1}{3} =m\]Ok good that gives us the `slope` of our tangent line.\[\Large y=mx+b \qquad\to\qquad y=\frac{1}{3}x+b\]To finish up the equation of our tangent line we'll plug in (0,0) to solve for our y-interecpt `b`.

Answer 11

b=0

Answer 12

Yay good job \c:/ we did it!
Here is a graph of the function and our tangent line in case you wanted to see it.
https://www.desmos.com/calculator/ya4viut3he

Answer 13

ok thanks! I learned this in the beginning of the semester and just never really practiced it so I sort of forgot it Thanks again!