Reversing Derivatives: Finding the Original Function and Point of Derivation

[skateza]

Homework Statement

Each of the limits represent a derivative f'(a). Find f(x) and a.
1) lim x-> 0 [(6^x) - 1]/x
2) lim x-> 0 [((4+h)^(-2)) - 0.0625]/h

2. The attempt at a solution

i'm not sure what I'm suppose to be solving for, they give us a derivative, or a partial derivative that's not completely solved and they want us to go backward to find the original function and the number that gave us the derivative they have. How do we do this without integrating.

 

[real10]

Homework Statement

Each of the limits represent a derivative f'(a). Find f(x) and a.
1) lim x-> 0 [(6^x) - 1]/x
2) lim x-> 0 [((4+h)^(-2)) - 0.0625]/h

2. The attempt at a solution

i'm not sure what I'm suppose to be solving for, they give us a derivative, or a partial derivative that's not completely solved and they want us to go backward to find the original function and the number that gave us the derivative they have. How do we do this without integrating.

How about using the definition of a derivative...

 

[skateza]

Okay but i never even thought of that...
lim x -> 0 [f(a+x)-f(a)]/x = lim x -> 0 [(6^x) - 1]/x

so does that mean f(a+x) = 6^x, and f(a) = 1

than to find f(x) you do f(a+x)-f(a) = 6^x - 1

than a = 6^a = 2 than you take the log of that to find a?

 

[HallsofIvy]

Okay but i never even thought of that...
lim x -> 0 [f(a+x)-f(a)]/x = lim x -> 0 [(6^x) - 1]/x

so does that mean f(a+x) = 6^x, and f(a) = 1

than to find f(x) you do f(a+x)-f(a) = 6^x - 1

than a = 6^a = 2 than you take the log of that to find a?

You never even thought of that? The problem said each was a derivative? Don't you associate derivatives with limits?

No, a is not 6^a and certainly not 2 (where did you get the 2 from?). What is 6^0?
Now what do you think f(x) and a are?

For the second problem you might want to calculate (1/4)² as a decimal number.

 

[skateza]

i was being sarcastic, of course that's the first thing i did...

i think on the bottom of my last post i meant to put

f(a+x) = 6^x, and f(a) = 1 Therefore, f(x) = 6^x - 1
if you stuff a in for x, to get f(a), you get it f(a) = 6^x - 1
Since f(a) = 1
1 = 6^x -1
6^x = 2
do a log to find a

OR... do i take the limit somewhere in there to find 6^0 = 1 and it'll all make sense...

 

[CompuChip]

What about f(0 + x) = 6^x (suggestive notation... hope that rings a bell)