AP Calc Easy Limits Question- I don't know what I'm doing wrong!?

Question

Find the slope of the curve f(x)= 1/(x-2) at x=3.

I know I'm supposed to use the equation (f(a+h)-f(a))/h, but somehow I just can't arrive at an answer. I think it has something to do with distributing my negative signs....

Thanks!

Answer 1

Since x = 3, we need to find [f(3+h) -f(3)]/h

Let's take this one step at a time.

f(3+h) = 1/(3+h-2) = 1/(h+1)

f(3) = 1/(3-2) = 1

Then f(3+h)-f(3) = 1/(h+1) - 1

Get a common denominator.

f(3+h)-f(3) = 1/(h+1) - (h+1)/(h+1) = (1-h-1)/(h+1) = -h/(h+1)

Then [f(3+h)-f(3)]/h = [-h/(h+1)]/h = -1/(h+1)

Now take the limit as h goes to 0.

-1/(0+1)

= -1

The slope is -1.