Originally Posted by

**Drexel28** I'll give you one more hint and let you take one more stab.

Consider the $\displaystyle \varphi(x)$ above.

Now, $\displaystyle \varphi$ being the difference of two continuous real-valued functions is continuous. Also, $\displaystyle \varphi(0)=f(0)-0=f(0)$. Now, if $\displaystyle \varphi(0)=0$ we're done, right? So what can you assume?

Similarly $\displaystyle \varphi(1)=f(1)-1$. If it's zero we're done again, so we can assume not.

So now you should have two conditions which should immediately make you jump for joy at knowing the IVT, what are they?