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

Question

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

William Abdalla · Accepted Answer

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.

Ezekiel Alexander · Answer

undefined To verify that $ y = x^3 + x - 1 $ over the interval $[0,2]$ satisfies the hypotheses of the Mean Value Theorem, we need to check two conditions:

1. The function $ y = x^3 + x - 1 $ must be continuous on the closed interval $[0,2]$.
2. The function $ y = x^3 + x - 1 $ must be differentiable on the open interval $(0,2)$.

1. Continuity:
To check continuity, evaluate the function at the endpoints of the interval $[0,2]$, namely at $x = 0$ and $x = 2$.

$y(0) = (0)^3 + 0 - 1 = -1$ 
   $y(2) = (2)^3 + 2 - 1 = 9$

Since the function is a polynomial, it is continuous everywhere, including the interval $[0,2]$.

2. Differentiability:
To check differentiability, we need to verify that the derivative of the function exists for all points in the open interval $(0,2)$.

$y'(x) = 3x^2 + 1$

The derivative exists for all real numbers, including the interval $(0,2)$.

Since both conditions are satisfied, the function $ y = x^3 + x - 1 $ over the interval $[0,2]$ satisfies the hypotheses of the Mean Value Theorem.