Question

Calculate the slope of the tangent to the given function at the given point.
a) f(x) = 2/ sqrt(x+5), p(4, 2/3)
b) f(x)= 3/(x+1), P(2,1)

we're supposed to use slope= [f(x+h)-f(x)]/h
whenever I use it, I keep getting zero for both, which can't be right.

Answer 1

you're right about that. So, how did you get zero? Don't just come around whining. Show us some of your work.

Let's start with f(x) = 1/√(x+5). Yeah, I know there's a 2 there, but that's just noise. Toss it back in at the end.

f(x+h) = 1/√(x+h+5)
f(x) = 1/√(x+5)
So, f(x+h) - f(x) = (√(x+5) - √(x+h+5))/(√(x+5)√(x+h+5))
Looks nasty, and it is. But if we multiply top and bottom by the "conjugate" (√(x+5) + √(x+h+5)) we get (since (a-b)(a+b) = a^2-b^2)
((x+5)-(x+h+5)) / (√(x+5)√(x+h+5)(√(x+5) + √(x+h+5)))
-h / / (√(x+5)√(x+h+5)(√(x+5) + √(x+h+5)))
Now divide by h and take the limit, and you get
-1 / 2(x+5)^(3/2)

Answer 2

do the other in the same way; it's a bit less messy. What do you get? How do you get it?

Answer 3

For the first one, I got -1/108. Would that be correct?