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  • Fill in the blanks in each of the An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is _______.

Fill in the blanks in each of the
An example of a function which is continuous everywhere but fails to be differentiable exactly at two points is _______.

Answers (1)

Consider, f(x) = |x-1| + |x-2|

Let’s discuss the continuity of f(x).

We have, f(x) = |x-1| + |x-2|

\begin{array}{l} f(x)=\left\{\begin{array}{l} -(x-1)-(x-2), \text { if } x<1 \\ (x-1)-(x-2), \text { if } 1 \leq x<2 \\ (x-1)+(x-2), \text { if } x \geq 2 \end{array}\right. \\ f(x)=\left\{\begin{array}{l} -2 x+3, \text { if } x<1 \\ 1, \text { if } 1 \leq x<2 \\ 2 x-3, \text { if } x \geq 2 \end{array}\right. \end{array}

When x<1, we have f(x) = -2x+3, which is a polynomial function and polynomial function is continuous everywhere.

When 1≤x<2, we have f(x) = 1, which is a constant function and constant function is continuous everywhere.

When x≥2, we have f(x) = 2x-3, which is a polynomial function and polynomial function is continuous everywhere.

Hence, f(x) = |x-1| + |x-2| is continuous everywhere.

Let’s discuss the differentiability of f(x) at x=1 and x=2.

We have

\begin{array}{l} f(x)=\left\{\begin{array}{l} -(x-1)-(x-2), \text { if } x<1 \\ (x-1)-(x-2), \text { if } 1 \leq x<2 \\ (x-1)+(x-2), \text { if } x \geq 2 \end{array}\right. \\ f(x)=\left\{\begin{array}{l} -2 x+3, \text { if } x<1 \\ 1, \text { if } 1 \leq x<2 \\ 2 x-3, \text { if } x \geq 2 \end{array}\right. \end{array}

\\\qquad LHD =\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ =\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{(1-h)-1} \\ =\lim _{h \rightarrow 0} \frac{[(-2(1-h)+3)-(-2+3)]}{(1-h)-1} \quad(\because f(x)=-2 x+3, \text { if } x<1)

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{[-2+2 h+3-1]}{-h}\\ &\lim _{=h \rightarrow 0} \frac{[2 h]}{-h}=\lim _{h \rightarrow 0} 2=2\\ &RLD =\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}\\ &=\lim _{x \rightarrow 1^{+}} \frac{1-1}{1-1}(\because f(x)=1, \text { if } 1 \leq x<2)\\ &=0\\ &\Rightarrow\left\lfloor f^{\prime}(1) \neq \operatorname{Rf}^{\prime}(1)\right.\\ &\Rightarrow f(x) \text { is not differentiable at } x=1 \text { . }\\ &L{ f^{\prime}(2)}=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2} \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 2} \frac{1-1}{2-2}(: f(x)=1, \text { if } 1 \leq x<2 \text { and } f(2)=2 \times 2-3=1)\\ &=0\\ &{R f^{\prime}(2)}=\lim_{x \rightarrow 2^{+} }\frac{f(x)-f(2)}{x-2}\\ &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(2)}{(1+h)-2}\\ &=\lim _{h \rightarrow 0} \frac{[(2(1+h)-3)-(2 \times 2-3)]}{(1+h)-2} \quad(\because f(x)=2 x-3, \text { if } x \geq 2)\\ &=\lim _{h \rightarrow 0} \frac{[2+2 h-3-1]}{h-1}\\ &=\lim _{h \rightarrow 0} \frac{[2 h-2]}{h-1}=\lim _{h \rightarrow 0} \frac{2(h-1)}{h-1}=2\\ &\Rightarrow \operatorname{Lf}^{\prime}(2) \neq \mathrm{Rf}^{\prime}(2) \end{aligned}

⇒ f(x) is not differentiable at x=2.

Thus, f(x) = |x-1| + |X-2| is continuous everywhere but fails to be differentiable exactly at two points x=1 and x=2.

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