Given the function f(x) defined by:
![f(x)=3-|x-3|,\text{ for }x\in\lbrack0,6]](https://img.qammunity.org/2023/formulas/mathematics/college/gnot9cnjn4mr6y2ha7mlj060bv40xwhwfx.png)
This function is continuous in [0, 6] and is differentiable everywhere except at the point x=3. This point is in the interval [0, 6], and since Rolle's Theorem requires that the function must be differentiable on the open interval (0, 6), we conclude that we can not use the Rolle's Theorem.