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  • Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.

Show that f(x) = |x - 5| is continuous but not differentiable at x = 5.

Answers (1)

Given,

f(x) = |x - 5| …(1)

We need to prove that f(x) is continuous but not differentiable at x = 5

A function f(x) is said to be continuous at x = c if,
Left hand limit(LHL at x = c) = Right hand limit(RHL at x = c) = f(c).
Mathematically we can represent it as-

\lim _{h \rightarrow 0} f(c-h)=\lim _{h \rightarrow 0} f(c+h)=f(c)

Where h is a very small number very close to 0 (h→0)

And a function is said to be differentiable at x = c if it is continuous there and

Left hand derivative(LHD at x = c) = Right hand derivative(RHD at x = c) = f(c).

Mathematically we can represent it as-
\\\lim _{x \rightarrow c^{-}} \frac{f(x)-f(c)}{x-c}=\lim _{x \rightarrow c^{+}} \frac{f(x)-f(c)}{x-c} \\\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{c-h-c}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{c+h-c}
Finally, we can state that for a function to be differentiable at x = c
\lim _{h \rightarrow 0} \frac{f(c-h)-f(c)}{-h}=\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}

Checking for the continuity:

Now according to above theory-

f(x) is continuous at x = 5 if -

\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)=f(5) \\ \therefore L H L=h \rightarrow 0 \\ \Rightarrow L H L=\lim _{h \rightarrow 0}|5-h-5|

\begin{aligned} &\Rightarrow \mathrm{LHL}=\lim _{\mathrm{h} \rightarrow 0}|-\mathrm{h}|\\ &\therefore \mathrm{LHL}=|-0|=0\\ &\therefore \mathrm{LHL}=0 \ldots(2)\\ &\text { Similarly, }\\ &\lim _{\mathrm{RHL}}=\lim _{h \rightarrow 0} \mathrm{f}(5+\mathrm{h})\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|5+\mathrm{h}-5|_{\{\text {using equation } 1\}}\\ &\Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}|\mathrm{~h}|\\ &\therefore \mathrm{RHL}=|0|=0 \ldots(3) \end{aligned}

And, f(5) = |5-5| = 0 {using equation 1} …(4)

From equation 2,3 and 4 we observe that:

\lim _{h \rightarrow 0} f(5-h)=\lim _{h \rightarrow 0} f(5+h)=f(5) =0

∴ f(x) is continuous at x = 5. So we will proceed now to check the differentiability.

Checking for the differentiability:

Now according to above theory

f(x) is differentiable at x = 2 if -

\begin{array}{l} \lim _{h \rightarrow 0} \frac{f(5-h)-f(5)}{-h}=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h} \\ \therefore L H D=h \rightarrow 0 \quad \frac{|5-h-5|-0}{-h} \\ \Rightarrow L H D=h \rightarrow 0 \frac{|-h|}{-h}\{\text { using equation } 1\} \end{array}

As h > 0 as defined in theory above.

∴ |-h| = h

\Rightarrow \mathrm{LHD}=\lim _{h \rightarrow 0} \frac{\mathrm{h}}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1 \\\therefore \mathrm{LHD}=-1 . .(5)
Now,
\\\operatorname{RHD}=\lim _{h \rightarrow 0} \frac{\mathrm{f}(5+h)-\mathrm{f}(5)}{\mathrm{h}} \\\Rightarrow \lim _{h \rightarrow 0} \frac{|5+h-5|-0}{h}=\lim _{h \rightarrow 0} \frac{|h|}{h}\{u sin g \ equation \ 1\}
As h>0 as defined in theory above.
\\\therefore|h|=h \\ \Rightarrow \lim _{\mathrm{RHD}}=\operatorname{h} \rightarrow 0^{\frac{\mathrm{h}}{\mathrm{h}}}=\lim _{\mathrm{h} \rightarrow 0} 1$ \\$\therefore \mathrm{RHD}=1 \ldots(6)

Clearly from equation 5 and 6,we can conclude that-

(LHD at x=5) ≠ (RHD at x = 5)

∴ f(x) is not differentiable at x = 5 but continuous at x = 5.

Hence proved.

Posted by

infoexpert22

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