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

Step2: Check continuity of \(f(x)=|x - 1|\) on \([0,2]\)

The function \(y = |x - 1|=

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

\). The limit as \(x\to1^{-}\) of \(f(x)\) is \(\lim_{x\to1^{-}}(1 - x)=0\), and the limit as \(x\to1^{+}\) of \(f(x)\) is \(\lim_{x\to1^{+}}(x - 1)=0\), and \(f(1)=0\). Also, \(f(x)\) is a linear - piecewise function, so it is continuous on \([0,2]\).

Step3: Check differentiability of \(f(x)=|x - 1|\) on \((0,2)\)

The derivative of \(y = 1 - x\) for \(x<1\) is \(y^{\prime}=- 1\), and the derivative of \(y=x - 1\) for \(x>1\) is \(y^{\prime}=1\). The left - hand derivative at \(x = 1\) is \(\lim_{h\to0^{-}}\frac{f(1 + h)-f(1)}{h}=\lim_{h\to0^{-}}\frac{1-(1 + h)-0}{h}=-1\), and the right - hand derivative at \(x = 1\) is \(\lim_{h\to0^{+}}\frac{f(1 + h)-f(1)}{h}=\lim_{h\to0^{+}}\frac{(1 + h)-1-0}{h}=1\). Since the left - hand derivative and the right - hand derivative at \(x = 1\) are not equal, \(f(x)\) is not differentiable at \(x = 1\in(0,2)\).

Answer:

No, \(f(x)=|x - 1|\) does not satisfy the Mean - Value Theorem on \([0,2]\) because it is not differentiable on the open interval \((0,2)\) (specifically, it is not differentiable at \(x = 1\)).