Question

I'm trying to understand:

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.

Does g(x) refer to an arbitrary function or is there something more to g(x) than I am aware?

Answer 1

It sounds like it should be arbitrary.

Answer 2

Yes, \(g(x)\) is an arbitrary function. The only thing we know about \(g\) is that it's continuous, which means that \[\lim_{x\rightarrow x_0} g(x) = g(x_0)\]for any value \(x_0\). That and the definition \(f(x) = (x-a)g(x)\) is all we know (and all we need to know) in order to solve the problem.

Since the problem is asking about \(f'(a)\) we use the formal definition of the derivative of \(f\) at \(a\):
\[f'(a) = \lim_{x\rightarrow a} \frac{f(x) - f(a)}{x-a}\] If we substitute the definition of \(f\) in terms of \(g\) this becomes
\[f'(a) = \lim_{x\rightarrow a} \frac{(x-a)g(x) - (a-a)g(a)}{x-a}\]The right hand term in the numerator is 0 (since \(a-a = 0\)) so this becomes
\[f'(a) = \lim_{x\rightarrow a} \frac{(x-a)g(x)}{x-a}\]In taking the limit the term \(x-a\) approaches 0 but never actually equals 0, so we can divide the \(x-a\) in the numerator by the \(x-a\) in the denominator to cancel those terms out. We then have
\[f'(a) = \lim_{x\rightarrow a} g(x)\] But because the function \(g\) is continuous we know that \(\lim_{x\rightarrow a} g(x)\) actually exists, and that \(\lim_{x\rightarrow a} g(x) = g(a)\). So finally we have
\[f'(a) = g(a)\] which is what we were asked to prove.

Answer 3

Oh, one neat thing I forgot add: This shows that the function \(f\) is differentiable at \(a\). But it's not necessarily the case that the function \(g\) is differentiable at \(a\): A function can be continuous but not necessarily (always) differentiable. For example, \(g\) could look as follows:

Here \(g\) is continuous at \(a\) but not differentiable at \(a\).