Rolle's theorem works for a function
over an interval
if:
is continuous on
![[a,b]](https://img.qammunity.org/2020/formulas/mathematics/middle-school/x9eybysx3cihgn4xxsxk4fl3z5dcnyp7ww.png)
is differentiable on
![(a,b)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/flm9yg0lwjqdn9jdt6jx1y4yg51gtx06uj.png)
![f(a)=f(b)](https://img.qammunity.org/2020/formulas/mathematics/college/g5rm1i51g958p2zz2sm8dtd8oa7lzdb7if.png)
This is our case:
is a polynomial, so it is continuous and differentiable everywhere, and thus in particular it is continuous and differentiable over [0,4].
Also, we have
![f(0)=7=f(4)](https://img.qammunity.org/2020/formulas/mathematics/college/jpdam72b8xb0wlh4kei5a2elm7bfms309r.png)
So, we're guaranteed that there exists at least one point
such that
.
Let's compute the derivative:
![f'(x)=3x^2-2x-12](https://img.qammunity.org/2020/formulas/mathematics/college/ppseuco2ganhdcm6a7q3jk4qeu3fbehit3.png)
And we have
![f'(x)=0 \iff x= (1\pm√(37))/(3)](https://img.qammunity.org/2020/formulas/mathematics/college/9mhcv10m4fdiv6c686kl6nfkt7u2i0vloa.png)
In particular, we have
![(1+√(37))/(3)\approx 2.36](https://img.qammunity.org/2020/formulas/mathematics/college/xidt8ttxpw420qk8x0l9njlnlerq3rxxzk.png)
so this is the point that satisfies Rolle's theorem.