Being a polynomial function, this function is certainly continuous and differentiable everywhere. In particular, it is continuous on the closed interval #[0,2]# and differentiable on the open interval #(0,2)#. This is enough to say it satisfies the hypotheses of the Mean Value Theorem.

Because of this, it also satisfies the conclusion of the Mean Value Theorem. There is a number #c\in (0,2)# with the property that #f'(c)=(f(2)-f(0))/(2-0)#. Since #(f(2)-f(0))/(2-0)=(9-(-1))/2=5#, this means there is a number #c\in (0,2)# such that #f'(c)=5#.

Since #f'(x)=3x^2+1#, we can find #c# by setting #3x^2+1=5# and solving for its positive root: #3x^2=4\Rightarrow x^2=4/3\Rightarrow x=2/sqrt(3)#. In other words, #c=2/sqrt(3)\approx 1.1547#. Finding #c# is not the point of the theorem (other than saying it's in the interval #(0,2)#), but it's nice to see that we can find it here anyway.