15. 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)).