Given $f(x) = 3x^4-2x^3-x^2 +1, \ g(x) = 4x^3-3x^2-2x$ show $\frac{f(1)-f(0)}{g(1) - g(0)} \ne \frac {f'(x)}{g'(x)}$ for any $x \in (0,1)$

Question

Given $f(x) = 3x^4-2x^3-x^2 +1, \ g(x) = 4x^3-3x^2-2x$

show $\frac{f(1)-f(0)}{g(1) - g(0)} \ne \frac {f'(x)}{g'(x)}$ for any $x \in (0,1)$

I'm a little confused. I did the fraction on the left and my result was $\frac{0}{-1}$ ... but I'm not sure this is what I should've gotten? Would limits make sense for the LHS?

The fraction on the RHS, $\frac{f'(x)}{g'(x)}=\frac{12x^3-6x^2-2x}{12x^2-6x-2}=x$.

Answer 1

You made a small mistake in the $g'(x)$. It should be $12x^2$ instead of $12x^3$ and as such, $\frac{f'(x)}{g'(x)} = x$

Answer 2

You obtained $0$ for the LHS and the term $x$ on the RHS. If $x$ is allowed to range over the open interval $\>]0,1[\>$ then it is never $=0$.

The idea of this neat example is the following: In the proof of Hôpital's rule we need the following tuned up version of the MVT: Under the given assumptions there is a $\xi\in\>]a,b[\>$ such that $${f(b)-f(a)\over g(b)-g(a)}={f'(\xi)\over g'(\xi)}\ .$$ This seems violated here. But an essential part of the assumptions is that $g'(x)\ne0$ for all $x\in\>]a,b[\>$, and this is violated in your example.