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

[canucks81]

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

 

[willem2]

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.

 

[AnTiFreeze3]

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.

 

[canucks81]

I did the question over again and I got -1/4 this time.

 

[AnTiFreeze3]

From what I solved, that's not the correct answer. But at least your answer was negative this time

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.

 

[Mark44]

Yes, show us what you did to get the value you show.

 

[canucks81]

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.

 

[Mark44]

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.

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.

 

[Mark44]

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.

 

[canucks81]

so using [f(a+h) - f(a) ] / h

is this the correct set up? [(1/(2+h)^2 - 1/16)] (1/h)

 

[Mark44]

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.

 

[Mark44]

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.