Problem

Each limit represents the derivative of some func…

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Problem 38 Easy Difficulty

Each limit represents the derivative of some function $ f $ at some number $ a $. State such an $ f $ and $ a $ in each case.

$ \displaystyle \lim_{h \to 0} \frac{e^{-2 + h} - e^{-2}}{h} $

Answer

$f(x)=e^{x}$
$a=-2$

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Video Transcript

we need to find a function F of X. And a number for a. Such that F prime the derivative of F F A is equal to this limit. Now the for function F of X and a value A. So X equals some number A The derivative of our function when X is A. Is equal to the limit of death of A plus H minus F of a. All divided by H. As we take the limit As a church approaches zero. So if we have some function F of X and some number a F prime of a equals the limit of F. Of a plus H minus F. Of a. Over H. As H goes towards era, we need to find, excuse me, we need to find a function F. Of X and a number A. Uh so that when we find F prime at a using the definition of the derivative, we will get this uh This limit for our function F. Of X. And for our number of a. Well, let's start comparing this to this. A plus H function of A plus H. Here we have E to the negative two plus H. A plus H negative two plus H. That's starting to make me think that maybe A is negative two. Now uh Here to function might be a little bit easier to see. Um Because you see this E here and maybe you think of E to the X. So let's write down uh possibly A is the number negative too? Possibly our function ffx is E to the X. Now if it turns out that ffx really is E. To the X and A equals negative two. Uh Then we're done because this is what we're actually supposed to be finding function F. Of X. And a number a such debt F prime at A equals this limit. Well, using F of X equals E. T. D. X. And using a equal to negative two. Let's use the definition for the derivative F. Prime of a equals this uh to see um if these are indeed uh the correct Effa Becks and correct. A. So we're going to apply this definition of the derivative for this function and this value of A. So if F of X equals E. To the X and A is negative two. F prime at A F. Prime at -2 would equal The limit as h approaches zero of F. Of A plus H. Now we're letting A B negative two. We're letting F of X. B. E. To the X. So F of -2 Plus H. F. F X. E. To the X. So F of negative two plus H would be E to the negative two plus H. Okay, our function is E. T. D. X. Our number for a is -2. So A plus H is negative two plus H. That's what we're plugging into the F of X. Function into the E. To the X. So F. Of A plus H. F of negative two plus H. Since F of X is E To the X. F of negative two plus H is going to be E to the negative two plus H. So we're looking pretty good because that's exactly what we have here. Now we have to subtract. Now we have to find F. Of A while A. Is negative two. F F X is E to the X. So F of negative two is E. To the negative too. And then that all gets put over H. And so if F of X equals E. T. D. X and A equals negative two, then F prime at negative two will equal this limit. Well, this limit expression is exactly what we started with and so F of X is E. T X. A is equal to negative two. And this limit expression is F prime of negative two. F of X equals E. To the X A is -2. F prime of a. F prime of negative two. Is this limit? So we did find the correct function and the correct value of it.

Leon D.

Temple University

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 2

Limits and Derivatives

Section 7

Derivatives and Rates of Change

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