(Sort of) Intermediate Value Theorem

[Yagoda]

Homework Statement

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).

Homework Equations

f'_{+}(a) = \lim_{x \rightarrow a^+}\frac{f(x)-f(a)}{x-a}

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.

 

[Office_Shredder]

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

 

[pasmith]

Homework Statement

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).

Homework Equations

f'_{+}(a) = \lim_{x \rightarrow a^+}\frac{f(x)-f(a)}{x-a}

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.

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.