# Problem on Mean Value Theorem

The Question:

Let $$f$$ be a function defined on an interval $$[a,b]$$. What conditions could you place on $$f$$ to guarantee that

$$\min f'\leq \frac{f(b)-f(a)}{b-a} \leq \max f'$$

We must require a $$\min f'$$ and a $$\max f'$$ before we can proceed further. That means $$f'$$ must be defined in $$[a,b]$$, that is, $$f$$ should be differentiable in $$[a,b]$$. So that's one condition.

Now, if $$f$$ is continuous in $$[a,b]$$ and differentiable in $$(a,b)$$ then, per the mean value theorem $$\exists \text{ } c\in(a,b) \text{ | } f'(c)=\frac{f(b)-f(a)}{b-a}$$

and obviously $$\min f'\leq f'(c) \leq \max f'$$

So the only condition would be for $$f$$ to be differentiable in $$[a,b]$$.

Now, I'll tell what's bothering me. The textbook I'm reading currently has the following definition for absolute extrema.

Let $$f$$ be a function with domain $$D$$. Then $$f$$ has an absolute maximum value on $$D$$ at a point $$c$$ if $$f(x) \leq f(c) \text{ } \forall \text{ } x \in D$$

and an absolute minimum value on $$D$$ at $$c$$ if $$f(x) \geq f(c) \text{ } \forall \text{ } x \in D$$

The reason, I've considered $$f$$ to be differentiable in $$[a,b]$$ and not just $$(a,b)$$ is because of the definition of absolute extrema.

Did I miss anything?

• Looks good to me.
– Zuy
Commented Sep 3, 2021 at 6:50
• A function can be differentiable on $[a,b}$, but its derivative needs not be bounded. Commented Sep 3, 2021 at 7:20
• @egreg That is exactly the doubt I'm having. The textbook that I'm reading right now, where I came across this problem has the following definition for extrema: If $f$ is a function with domain $D$, then $f$ has an absolute extremum value on $D$ at a point $c$ if $f(x)\leq f(c)$ or $f(x)\geq f(c)$ (as applicable) for all $x$ in $D$. That is why I considered $f'$ to be bounded so that it can have extrema, though, I can make up functions that are unbounded and have extrema. I'm confused. Commented Sep 3, 2021 at 7:30
• @AbhishekAUdupa If a function is upper and lower unbounded, it cannot have extrema. Commented Sep 3, 2021 at 7:38
• @egreg so then, if $f'$ is unbounded then $\max f'$ and $\min f'$ wouldn't exist right? Commented Sep 3, 2021 at 7:43

You need to assume that the derivative is bounded if you want to actually bound the ratios $$\frac{f(b)−f(a)}{b−a}$$.
If you look at $$f$$ defined as:
• $$f(x) = \sin(1/x^2)\times x^2$$ if $$x\in(0,1]$$
• $$f(0) = 0$$
then $$f$$ is differentiable over all $$[0,1]$$, but its derivative is not bounded, as for $$x>0$$:
$$f'(x)=2x\sin\left(\frac1{x^2}\right) -\frac2x\times\cos\left(\frac1{x^2}\right)$$
and since the derivative values are limits of ratios $$\frac{f(b)−f(a)}{b−a}$$, those ratios are not bounded either.