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Function with Given Properties – Problem of the Week

Function with Given Properties – Calculus Problem of the Week

This Calculus Problem of the Week exemplifies a type of college-level exam problem that appears from time to time, and involves the definition of the derivative. It is unlike most homework problems that you will have done. (You know how some teachers love to throw a new type of problem on exams to “test how well you really understand the material”?!?) We don’t think high-pressure test situations are actually a good way to test your understanding through such problems, and instead mostly measure how well you respond to stress, a totally different thing. We thus believe it’s helpful to have worked through such a problem before you see it on an exam.

So let’s do that now!

Practice Exam Problem: Function with Given Properties

Let $f(x)$ be a function defined for all real values of x, and having the following properties:

i. $f(a+b) = f(a) \cdot f(b)$ for all real numbers a and b.
ii. $f(0) = 1$
iii. $f(x)$ has a derivative at $x = 0.$

Show that $f(x)$ has a derivative for all real values of x, and that $f'(x) = f'(0) \cdot f(x).$


This problem may seem intimidating at first, which is why we’re glad you’re seeing it now. We’ll not only provide the solution, but also some problem-solving tips to help you when you encounter a non-routine problem on an exam.

One key thing to know is that in a problem like this, you just have to dive in, try some things, and see what that gets you. Don’t stare at the blank page and wait for inspiration; instead, take a first step in any direction that strikes you after reading whatever the question asks you to do. If that doesn’t work, try something else. The key thing is to start.

Here, since the problem tells us first to “show that $f(x)$ has a derivative for all real values of x,” without having any place else to start we begin with the definition of the derivative. If in doubt, if a problem asks something general about a derivative, start with the definition! (That’s a very “mathematical” thing to do, which of course math professors love.) \[f'(x) = \lim_{h \to 0}\frac{f(x+h) – f(x)}{h}\]

Let’s use Property (i) to rewrite the first term in the numerator, and see what happens. \begin{align*} f'(x) &= \lim_{h \to 0}\frac{f(x+h) – f(x)}{h} \qquad [\text{Property (i): }f(a+b) = f(a) \cdot f(b)]\\[8px] &= \lim_{h \to 0} \frac{f(x) \cdot f(h) – f(x)}{h} \\[8px] &= \lim_{h \to 0} \frac{f(x) \big[f(h) – 1 \big]}{h} \\[8px] &= \lim_{h \to 0} f(x) \cdot \lim_{h \to 0}\frac{f(h) – 1}{h} \\[8px] &= f(x) \cdot \lim_{h \to 0}\frac{f(h) – 1}{h} \quad \blacktriangleleft \end{align*} Note that $\displaystyle{\lim_{h \to 0}f(x) = f(x)}$ since there is no h-dependence in $f(x).$

Hmmm. It’s not immediately clear what to do next, but we remember that we’re trying to show that $f'(x) = f'(0) \cdot f(x).$

And oh, wait, we currently have from above \[f'(x) = f(x) \cdot \lim_{h \to 0}\frac{f(h) – 1}{h} \quad \blacktriangleleft\] So if we can show $\displaystyle{\lim_{h \to 0}\frac{f(h) – 1}{h} = f'(0)},$ we’ll be done.

We also remember that we have two properties we haven’t used yet: “(ii) $f(0) = 1$” and “(iii) $f(x)$ has a derivative at $x = 0.$” So let’s see if we can do some manipulation with those properties and convert our remaining limit into $f'(0).$

Subproblem to examine $f'(0)$

Let’s again use the definition of the derivative, now for $f'(0)$, and see what that gets us. We can do this because Property (iii) explicitly tells us that $f'(0)$ exists, so we can certainly apply the definition of the derivative at $x = 0.$ \[f'(0) = \lim_{h \to 0}\frac{f(0 + h) – f(0)}{h} \] Oh, but wait: Property (ii) says $f(0) = 1.$ So we have \[f'(0) = \lim_{h \to 0}\frac{f(h) – 1}{h} \] And that’s exactly what we were aiming to show! We didn’t see how that would fall out earlier, but lo and behold, there it is. (Whew!)

End subproblem.

So picking up from the line marked $\blacktriangleleft$ above, we have \begin{align*} f'(x) &= f(x) \cdot \lim_{h \to 0}\frac{f(h) – 1}{h} \quad\blacktriangleleft \quad [\text{Property (ii): }f(0) = 1]\\[8px] &= f(x) \cdot \lim_{h \to 0} \frac{f(0 + h) – f(0)}{h} \\[8px] &= f(x) \cdot f'(0) \end{align*}

So we’ve shown that \[f'(x) = f'(0) \cdot f(x) \quad \cmark\] Now we just have to write some words to show that $f(x)$ has a derivative for all real values of x:

We are told in the problem statement that $f(x)$ is defined for all real values of x, and Property (iii) states that $f(x)$ has a derivative at $x = 0$, $f'(0).$ Since we have shown that $f'(x) = f'(0) \cdot f(x),$ we conclude that $f'(x)$ exists for all $x. \; \cmark$

As we said, this problem is a little different than your typical homework question. What do you think? Is it easy for you? Difficult? Bewildering? Or do you have any other “unusual” problems that you’re working to learn to solve? Let us know in the comments below, or (better) on our Forum, and we’ll do our best to help.

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