Science:Math Exam Resources/Courses/MATH110/April 2011/Question 01 (b)/Solution 1

This statement is false. It reads like the conclusion of Rolle's Theorem; however, Rolle's Theorem has two conditions: that $f$ be continuous on a closed interval (in this case the interval $[0,1]$) and differentiable on the same open interval (in this case $(0,1)$). If we drop either of these two conditions, then the conclusion of Rolle's Theorem does not necessarily hold.

For example, suppose $f$ is not differentiable on $(0,1)$. Then we can draw the picture:

While this graph satisfies $f(0)=f(1)=1$ there is no point where the derivative is zero, because the derivative is either 1, -1, or undefined.

Suppose $f$ is not continuous on $[0,1]$. Then you can draw a picture like this:

Where clearly the derivative is never zero, even though $f(0)=f(1)=1$.

Either example (or a similar example) would suffice to prove the statement false.