How do you verify that y = x^3 + x - 1 over [0,2] satisfies the hypotheses of the Mean Value Theorem?

1 Answer
May 17, 2015

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.