Question

1. does $f(x)=|x - 1|$ satisfy the mean value theorem on $0,2$? explain.

Explanation:

Step1: Recall Mean - Value Theorem conditions

The Mean - Value Theorem states that if \(y = f(x)\) is continuous on the closed interval \([a,b]\) and differentiable on the open interval \((a,b)\), then there exists at least one number \(c\in(a,b)\) such that \(f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}\). Here, \(a = 0\), \(b = 2\), and \(f(x)=|x - 1|\).

Step2: Check continuity

The function \(y=|x - 1|\) can be written as \(f(x)=

$$\begin{cases}1 - x, &x<1\\x - 1, &x\geq1\end{cases}$$

\).
\(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}(1 - x)=0\), \(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}(x - 1)=0\), and \(f(1)=0\). So \(f(x)\) is continuous on \([0,2]\).

Step3: Check differentiability

The derivative of \(y = f(x)\) for \(x<1\) is \(f^{\prime}(x)=-1\), and for \(x > 1\) is \(f^{\prime}(x)=1\).
\(\lim_{x
ightarrow1^{-}}\frac{f(x)-f(1)}{x - 1}=\lim_{x
ightarrow1^{-}}\frac{1 - x-0}{x - 1}=-1\) and \(\lim_{x
ightarrow1^{+}}\frac{f(x)-f(1)}{x - 1}=\lim_{x
ightarrow1^{+}}\frac{x - 1-0}{x - 1}=1\).
The left - hand derivative and the right - hand derivative at \(x = 1\) are not equal. So \(f(x)\) is not differentiable on \((0,2)\).

Answer:

No, because \(f(x)=|x - 1|\) is not differentiable on the open interval \((0,2)\) although it is continuous on the closed interval \([0,2]\).