Solve problems of the form "given a differentiable function prove that there exists a point c such that"

Question

The problems in question all have essentially the same setup: Given a function $f$ that is continuous on $[a,b]$ and differentiable on $(a,b)$ prove that there exists a point $c \in (a,b)$ such that $f'(c) = ...$

Here are some examples:

1. $f: \mathbb{R} \rightarrow (0,\infty)$, $f$ differentiable and $f(1) = 1$. Prove that there exists $c$ such that $e^{2021\frac{f'(c)}{f(c)}} = f(2022)$.
2. $f$ is differentiable on $\mathbb{R}$, $a > 0$ and $f(a) = 0$. Prove that there exists $c$ such that $f(c) +2cf'(c) = 0$.
3. $f$ is continuous on $[0,1]$ and differentiable on $(0,1)$, $f(0)=0$. Prove that there exists $c \in (0,1)$ such that $f(1)(f(1)-1)) = f'(c)(2f(c)-1)$.
4. $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, $f(x) \neq 0$ for all $x \in (a,b)$. Prove that there exists $c$ such that $\frac{f'(c)}{f(c)} = \frac{1}{a-c} + \frac{1}{b-c}$.

In all these, the solution essentially boils down to defining a new function $F$ and applying Rolle's theorem or the mean value theorem on that function and some specific points. Is there any general way of solving these problems, or is there any better way of approaching these than just trying to guess the function or hoping to notice something specific?

Answer 1

You've already identified what I think is the best way to approach all of the problems - they're all quite clearly designed to make you think of Rolle's Theorem and/or the Mean Value Theorem in their setup.

From that point, you probably want to look at the expression you're working with and try to figure out how it invokes RT/MVT. That means you're looking for a differentiable function that is equal at two endpoints of an interval, and you want to use the fact that it has zero derivative to relate to the actual thing you're proving.

Some things that you can keep an eye out for include:

1. If you're only given one point to work with, but Rolle's Theorem needs two, then you probably need to force another point into your new function (e.g. if you can make $g(0) = 0$, that seems pretty useful in several of the questions).

2. Think of other transformations that make sense given the expression - when I see $\frac{f'(x)}{f(x)}$ my immediate thought is $g(x) = \ln f(x)$ because its derivative is exactly what we need. Similarly if you don't have $f(a) = f(b) = 0$, then you must need to do some kind of shift so that $g(a) = g(b) = 0$.