Chapter 4.2, Problem 45E

To determine

To explain: The reason why it does not contradict the Mean Value Theorem.

Answer to Problem 45E

Due to the non-continuous nature of this function, it does not violate the Mean Value Theorem. Differentiable and continuous functions are covered by the theorem.

Explanation of Solution

Given information:

The function is:

  $f(x)=\left\{\begin{array}{c}x,0\le x<1\\ 0,x=1\end{array}\right.$

Explain:

It is important to remember that the Mean Value Theorem states that if a function $y=f(x)$ is continuous at every point of the closed interval $(a,b)$ and differentiable at every point of its interior $(a,b)$ , then there is at least one point $c\in (a,b)$ at which is

  $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}$

The graph of the given function to see it is continuous on the interval $[0,1]$

The function is defined as $f(x)=x$ on the interval $[0,1]$ and it is $f(x)=0$ at

the point $x=1$

The graph of the function shown below:

  

See that the first need is that the function $\text{f}$ must be continuous on the closed interval by reaching the Mean Value Theorem.

This explains why the Mean Value Theorem is not violated by the function $\text{f(x)}$ .