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18.01SC Single Variable Calculus:
Problem Set 1.
Question 1C2:
Let f(x) = (x − a)g(x). Use the definition of the derivative to calculate that f'(a) = g(a), assuming that g is continuous.
I'm not understanding the solution.
Could someone explain it to me step by step?
I know what the definition of a derivative is. I'm assuming "x" and "a" are two different points on the graph?
Any help would be appreciated!!
 8 months ago
 8 months ago
18.01SC Single Variable Calculus: Problem Set 1. Question 1C2: Let f(x) = (x − a)g(x). Use the definition of the derivative to calculate that f'(a) = g(a), assuming that g is continuous. I'm not understanding the solution. Could someone explain it to me step by step? I know what the definition of a derivative is. I'm assuming "x" and "a" are two different points on the graph? Any help would be appreciated!!
 8 months ago
 8 months ago

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choij1sBest ResponseYou've already chosen the best response.0
I guess I'm also not understanding the question. It's hard for me to picture it visually.
 8 months ago

choij1sBest ResponseYou've already chosen the best response.0
Could I say that the question is asking me to find the lim of f(x) as "x" approaches "a" ?
 8 months ago

creeksiderBest ResponseYou've already chosen the best response.1
In this problem, "x" is the independent variable, "a" is an unspecified constant, and "g(x)" is an unspecified function of x. It may be easier to visualize this if we assign values, so that a=3 and g(x)=sin x. Then we would have\[f(x)=(x3)\sin x\]and you would be trying to show that f'(3) =sin 3. The problem asks you to show the more generalized answer that works for any constant "a" and any function "g(x)". You aren't trying to find the limit of f(x). You're trying to find the limit of the difference quotient, which is used in the definition of the derivative. We're used to seeing the difference quotient looking like this:\[\frac{ f(x+\Delta x)f(x) }{ \Delta x }\]In this case we're interested in the derivative at a specific value of x, namely "a", so you see a different form of the difference quotient being used in the solution. I don't recall whether Prof. Jerison used this version in class, but it works as indicated in the solution provided.
 8 months ago

choij1sBest ResponseYou've already chosen the best response.0
Thanks for a reply. Two questions though; 1) am I finding the limit of g(x) as "x" approaches "a" then? 2) Why does f(a) turn into 0 in the solution?
 8 months ago

creeksiderBest ResponseYou've already chosen the best response.1
1) The first part of the solution establishes that the difference quotient is equal to g(x), and once this is established, the remaining part of the solution is the fairly trivial observation that g(x) goes to g(a) as x goes to a. Note that there is one nontrivial aspect of this step: we can't say this unless we know g is continuous, and that is why the exercise makes a point of setting up this fact. 2) In this problem, f(x) is (xa)g(x), so f(a) is (aa)g(a), which is 0*g(a), which is 0.
 8 months ago

choij1sBest ResponseYou've already chosen the best response.0
Oh okay. Thanks a lot!!
 8 months ago
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