# Thread: Did I do this problem correctly (equation of the line tangent)

1. ## Did I do this problem correctly (equation of the line tangent)

the problem goes as so...find the equation of the line tangent to f(x) = -1/x at the point (3, -1/3). Using the four-step process to find f'(x), I got:

f'(x) = 1/x^2

then using that derivative, i found the slope of the tangent line as f'(3) = 1/3^2 or 1/9.

thus, f(x) is f(3) = -1/3

lastly using y-y0=m(x-x0) i got y + 1/3 = 1/9(x-3)

solving that i got 1/9x -.6 (repeating)

does that sound right? thanks

2. Originally Posted by calcconfused
the problem goes as so...find the equation of the line tangent to f(x) = -1/x at the point (3, -1/3). Using the four-step process to find f'(x), I got:

f'(x) = 1/x^2

then using that derivative, i found the slope of the tangent line as f'(3) = 1/3^2 or 1/9.

thus, f(x) is f(3) = -1/3

lastly using y-y0=m(x-x0) i got y + 1/3 = 1/9(x-3)

solving that i got y=1/9x -.6 (repeating)

does that sound right? thanks
Correct (be aware of the red addendum)

You could also write $\displaystyle -\frac 23$ instead of -0.6.....
it'd be better