Mean Value Theorem:
Let f(x)=(x-3)^(-2). Show that there is no value of c in (1,4) such that f(4)-f(1)=f'(c)(4-1). Why does this not contradict the Mean Value Theorem?

Question

Mean Value Theorem:
Let f(x)=(x-3)^(-2).  Show that there is no value of c in (1,4) such that f(4)-f(1)=f'(c)(4-1).  Why does this not contradict the Mean Value Theorem?

anonymous · Answer

Where is $f(x)$ continuous?

anonymous · Answer

I think that's my problem - I don't know how to tell whether it's continuous or not unless it's graphed

anonymous · Answer

$$f(x)=(x-3)^{-2}=\frac{1}{(x-3)^2}$$
What would happen is $x=3$ ?

anonymous · Answer

Oh!  It wouldn't exist

anonymous · Answer

So rule of thumb is that if x is in a position for denominator to be zero then it is not continuous

anonymous · Answer

Yes, that's a way of thinking of it. A vertical asymptote occurs at $x=3$, if you were to look at the function's plot.

So why can't you apply the MVT?

anonymous · Answer

Because 3 is within (1,4) but it DNE

anonymous · Answer

Right. More specifically, a function has to be continuous over a given interval. This $f(x)$ is not, and so MVT does not apply.

anonymous · Answer

Perfect, thanks for your help!

anonymous · Answer

you're welcome