Find f '(2) if f(x) = 1/(x+2)^2

  • Thread starter canucks81
  • Start date
In summary: Check that your answer is correct. In summary, the homework equation is f(a+h) - f(a) = 0 when h --> 0.
  • #1
canucks81
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Homework Statement



Use the definition of derivative to Find f ' (2) if f(x) = 1/(x+2)^2

Homework Equations



lim [ f(a+h) - f(a) ]/ h
h->0

3. Attempt
The answer I get is 1/8. Can someone tell me if this is correct? Thanks
 
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  • #2
canucks81 said:

Homework Statement



Use the definition of derivative to Find f ' (2) if f(x) = 1/(x+2)^2

The answer I get is 1/8. Can someone tell me if this is correct? Thanks

No. f(x) = 1/(x+2)^2 is a decreasing function for x>0, so f'(2) can't be positive.
 
  • #3
I originally was going to post what I got, but that might be a little too revealing, so just know that my answer agrees with what Willem said, namely that the answer needs to be negative.

Finding derivatives using the definition of the derivative and the Newton Quotient can become very messy when you have to deal with all of these terms, so I would recommend doing the whole problem again, and if the problem persists, then you can just post your work here and someone, maybe me, should be able to find your mistake for you.
 
  • #4
I did the question over again and I got -1/4 this time.
 
  • #5
From what I solved, that's not the correct answer. But at least your answer was negative this time :smile:

Try working through it again, but this time, type out what you did (or take a picture of it and post that) so that, if you don't get the correct answer, someone can figure out where you're going wrong.
 
  • #6
Yes, show us what you did to get the value you show.
 
  • #7
lim x-> 2 (1/(x+2)^2) - 2 / (x-2)

lim x ->2 [ 1 - 2(x+2)^2 / (x+2)^2 ] / (x-2)

lim x->2 [ 1 - (2x^2 + 8x + 8 ) / (x+2)^2 ] / (x-2)

lim x->2 [ 1 -(2x+4)(x+2) / (x+2)^2 ] / (x-2)

lim x ->2 [ 1 - 2 (x+2)^2 / (x+2)^2 ] / (x-2 )

(x+2)^2 cancel out

lim x -> 2 -1/(x-2) = -1 / (2-2) = -1/0 = negative infinity

My mistake was that I used (x+2) instead of (x+2)^2 so I was just a little careless. I'm pretty sure this is the correct answer.
I still feel like my 3rd step is wrong. If i write it out as 1 - 2x^2 - 8x - 8 then this will eventually turn into -2x^2 - 8x - 7.
 
Last edited:
  • #8
canucks81 said:
lim x-> 2 (1/(x+2)^2) - 2 / (x-2)

lim x ->2 [ 1 - 2(x+2)^2 / (x+2)^2 ] / (x-2)

lim x->2 [ 1 - (2x^2 + 8x + 8 ) / (x+2)^2 ] / (x-2)

lim x->2 [ 1 -(2x+4)(x+2) / (x+2)^2 ] / (x-2)

lim x ->2 [ 1 - 2 (x+2)^2 / (x+2)^2 ] / (x-2 )

(x+2)^2 cancel out

lim x -> 2 -1/(x-2) = -1 / (2-2) = -1/0 = negative infinity

My mistake was that I used (x+2) instead of (x+2)^2 so I was just a little careless. I'm pretty sure this is the correct answer.
No, it's not.
canucks81 said:
I still feel like my 3rd step is wrong. If i write it out as 1 - 2x^2 - 8x - 8 then this will eventually turn into -2x^2 - 8x - 7.

As I set it up, I started with (1/h)[f(2 + h) - f(2)].

After doing the algebra, I then took the limit as h → 0, and got a number.

Your undefined value makes no sense. Your function is continuous everywhere except at x = -2, so it has a derivative everywhere but at x = -2.
 
  • #9
canucks81 said:
lim x-> 2 (1/(x+2)^2) - 2 / (x-2)
You started out wrong right off the bat. This is not the correct difference quotient for f(x) = 1/(x + 2)2. See my previous post for a better way to start.
 
  • #10
so using [f(a+h) - f(a) ] / h

is this the correct set up? [(1/(2+h)^2 - 1/16)] (1/h)
 
  • #11
canucks81 said:
so using [f(a+h) - f(a) ] / h

is this the correct set up? [(1/(2+h)^2 - 1/16)] (1/h)
No.

Since you're looking for f'(2), start with [f(2+h) - f(2) ] / h

Now, what is f(2 + h)? It's NOT 1/(2 + h)2.

f(2) = 1/16 is correct, though.
 
  • #12
If you do it this way, you won't have to drag lim h --> 0 along for so many steps.

1. Write [f(2+h) - f(2) ] / h using the definition of your function.
2. Simplify what you have from step 1.
3. Take the limit as h --> 0.
 

FAQ: Find f '(2) if f(x) = 1/(x+2)^2

What is the formula for finding f'(2)?

The formula for finding the derivative of a function at a specific point is f'(x) = lim(h->0) [f(x+h) - f(x)]/h. For this problem, we substitute x=2 into the given function to find f'(2).

How do I find the derivative of a fraction?

To find the derivative of a fraction, we use the power rule of differentiation. In this case, we have a fraction raised to the power of -2, so the derivative becomes -2(x+2)^-3. We then substitute x=2 into this expression to find f'(2).

What does f'(2) represent?

f'(2) represents the slope of the tangent line to the function f(x) at the point x=2. This is also known as the instantaneous rate of change of the function at x=2.

Why is it important to find f'(2)?

Finding f'(2) allows us to understand the behavior of the function at the specific point x=2. It can help us determine if the function is increasing or decreasing at that point and the steepness of the curve at that point.

Is there a shortcut for finding f'(2) without using the limit definition?

Yes, there is a shortcut called the power rule of differentiation. This rule states that the derivative of x^n is nx^(n-1), where n is any real number. In this case, we use the power rule to find the derivative of our given function, and then substitute x=2 into the resulting expression to find f'(2).

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