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Let $f(x)\leq g(x)$ for all $x \in I$, where $I$ is an interval $\subseteq$ R. Also, let $f(c) = g(c)$ for some $c \in I$ but not an endpoint. Prove that $f'(c) = g'(c)$ (assume differentiablity)

I have tried the Mean Value Theorem, let I = [a,b]. So for $f'(c) = g'(c)$ I will need to prove that $f(a) - f(b) = g(a) - g(b)$, but I am stuck at this. I have also tried the definition of derivative but I am still unable to go ahead.

Please give me some hints on this. Thank you!

sucksatmath
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2 Answers2

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If we take the function $h(x) = g(x) - f(x)$
we see that $h(x) \ge 0 = h(c)$ for all $x \in I$

So $c$ is a point of local minimum for $h$ is in the interval $I$.

Since $c$ is internal point to $I$, and $h$ is differentiable at $c$
(and that's so because both $f$ and $g$ are differentiable at $c$),
it follows that $h'(c) = 0$ (there is a very basic theorem stating this).

Fermat's theorem (stationary points)

This theorem is more basic than the mean value theorem (MVT),
meaning it's usually proven before the MVT is proven (in real analysis courses).

So now we have $0 = h'(c) = f'(c) - g'(c)$

So it follows that $f'(c) = g'(c)$

So all in all, your problem is just a simple exercise of this theorem.
I don't think the problem is MVT related.

If you're interested in a proof, it's very simple

Proof of Fermat's theorem

peter.petrov
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  • Hi, I have a question: can we extend this proof to the case when c is at the boundary (let's say the left one), can we prove that $f'(c) \leq g'(c)$ – sucksatmath Mar 26 '21 at 17:57
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    @sucksatmath Yes, to prove this inequality won't be a problem, it follows trivially by taking a limit. But if it's on the boundary then of course we just have right/left derivative there (at point c). Regarding the equality - I am not sure right now. – peter.petrov Mar 26 '21 at 19:34
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    The looser restriction with $c$ at the left boundary corresponds to the case in my answer with $x > c$. – Ben Mar 26 '21 at 20:32
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Here's another view, though @peter.petrov's excellent answer is better IMO and points towards a very useful generalization.

Let's look at the ratio $\frac{f(x) - f(c)}{x-c}$ for points near $c$.

By hypothesis for all $x$ in the interval $$f(x) - f(c) \leq g(x) - f(c) = g(x) - g(c)$$

for $x > c$ then, we have $$\frac{f(x) - f(c)}{x-c} \leq \frac{g(x) - g(c)}{x-c}$$

$$\lim_{x \to c+} \frac{f(x) - f(c)}{x-c}\leq \lim_{x \to c+} \frac{g(x) - g(c)}{x-c}$$ or $$f'(c) \leq g'(c)$$ However, for $x$ approaching $c$ from the left, the denominator $(x-c)$ is negative, so the inequality will be flipped, leaving us with

$$f'(c) \geq g'(c)$$

Taking these two statements together we have $$f'(c) \leq g'(c) \leq f'(c)$$ Therefore, $f'(c) = g'(c)$.

Ben
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