# (Sort of) Intermediate Value Theorem

1. Sep 8, 2013

### Yagoda

1. The problem statement, all variables and given/known data
Suppose that f is continuous on [a,b], its right-hand derivative $f'_{+}(a)$ exists, and $\mu$ is between $f'_{+}(a)$ and $\frac{f(b)-f(a)}{b-a}$. Show that $f(c) - f(a) = \mu (c-a)$ for some $c \in (a,b)$.

2. Relevant equations
$f'_{+}(a) = \lim_{x \rightarrow a^+}\frac{f(x)-f(a)}{x-a}$

3. The attempt at a solution
I spent a little while trying to figure out exactly what I am trying to prove and it definitely makes sense now. Intuitively I can see that it is true. My first instinct was to use the IVT for derivatives or the MVT, but those both require that f be differentiable, which I don't know to be the case here.
So I looked at using the regular IVT, since it only requires that f be continuous on [a,b], but also needs f(a) not equal to f(b). I was going to for now ignore the case where f(a) = f(b), but I was having trouble showing that $\mu$ is between the required bounds.
Am I even on the right track or is there another approach that makes more sense?

Thanks.

2. Sep 8, 2013

### Office_Shredder

Staff Emeritus
It might help to know that the function
$$\frac{ f(x) - f(a)}{x-a}$$
is a continuous function on (a,b)

3. Sep 8, 2013

### pasmith

The question suggests that it would be worth considering
$$g(x) = \frac{f(x) - f(a)}{x - a}$$
which is certainly defined and continuous for $a < x \leq b$. Since $f'_{+}(a)$ exists we can make $g$ continuous at $x = a$ by setting $g(a) = f'_{+}(a)$.

Now $g$ is continuous on $[a,b]$ and you can apply the IVT.