Asked by julion
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.
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.
Answers
Answered by
oobleck
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)
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)
Answered by
oobleck
do the other in the same way; it's a bit less messy. What do you get? How do you get it?
Answered by
A
For the first one, I got -1/108. Would that be correct?
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