If f(x) = 1/x, find f'(3) and use it to find the equation of the tangent to the curve y = 1/x at the point (3, 1/3).
I found the slope to be -1/9 but I don't know how to find the equation
You have done
$\displaystyle
f(x)= \frac{1}{x}
$
$\displaystyle
\frac{dy}{dx}= \frac{-1}{x^2}
$
And thus got answer $\displaystyle \frac{-1}{9}$
Now equation of a line ,passing through (a,b),with a slope "m" is given by
$\displaystyle
(y-b)=m(x-a)
$
So the answer is
$\displaystyle
(y-\frac{1}{3}) = \frac{-1}{9} (x-3)
$