So I was doing some calculus homework the other day, and the following question occurred to me: what functions have the property that the value of $c$ guaranteed by the Mean Value Theorem for Derivatives is the same as the value of $c$ guaranteed by the Mean Value Theorem for Integrals? By scaling, it's sufficient to consider $(0,1)$, so the condition is

$$f'(c)=f(1)-f(0)\,\,\,\, \text{and}\,\,\,\, f(c)=\int_0^1 f(x)\, dx.$$

After some experimentation, I found that $f(x)=e^x$ works, but I was wondering if there are other solutions, and if not, how you could go about showing that no other solutions exist.

I'm fairly certain this is the only solution, since MVT for Derivatives is just MVT for Integrals applied to $f'$, and $e^x$ is the solution to $f=f'$, but I was wondering if anything "weird" happened to give other solutions.


1 Answer 1


The mean value theorem promises that a function will at some point, call it $c\in X$ the domain, attain the average rate of change over an interval. The mean value theorem for integration promises that a function will at some point attain the average hight over the interval.

if $F'(x)=f(x)$ is an anti-derivative of $f(x)$...

$f(c) = \frac{\int_a^b f(x)dx}{b-a}$

$f(c) = \frac{F(b) - F(a)}{b-a} = V$

We have the average value, V.

Now take the derivative of both sides. $f'(c) = \frac{1}{b-a}[F'(b)-F'(a)]$

$f'(c) = \frac{1}{b-a}[f(b) -f(a)] = W$ and that is the average rate of change.

but, $e^x$ is not the only solution.

We need c such that:

$f(c) = V$ a particular constant.

and c

$f'(c) = W$ a different constant.

that's where I stared


$f(x) = a_0x+a_1$

on the interval a to b we want

$f(c) = \frac{a_0b^2 +a_1b-a_0a^2 -a_1a}{b-a}$ now we find c

$a_0c+a_1 = \frac{a_0(b -a)(b+a) +a_1(b- a)}{b-a}$

$a_0c= a_0b+a_0a$

$c=b+a$ very nice.

and $f'(x) = a_0$, $\forall x$

hence we have the correct slope.

  • $\begingroup$ Okay so linear functions work too. Anything else? Is this the complete solution set? $\endgroup$
    – Samir Khan
    Apr 12, 2015 at 3:40
  • $\begingroup$ there is a constant function that works... $\endgroup$
    – futurebird
    Apr 12, 2015 at 4:50

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