15. does $f(x)=|x - 1|$ satisfy the mean value theorem on $0,2$? explain.
15. does $f(x)=|x - 1|$ satisfy the mean value theorem on $0,2$? explain.
Answer
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\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}(1 - x)=0), (\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}(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\rightarrow1^{-}}\frac{f(x)-f(1)}{x - 1}=\lim_{x\rightarrow1^{-}}\frac{1 - x-0}{x - 1}=-1) and (\lim_{x\rightarrow1^{+}}\frac{f(x)-f(1)}{x - 1}=\lim_{x\rightarrow1^{+}}\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]).