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  • Examine the differentiability of f, where f is defined by f(x) = 1+x

Examine the differentiability of f, where f is defined by
f(x)=\left\{\begin{array}{ll} 1+x, & \text { if } x \leq 2 \\ 5-x, & \text { if } x>2 \end{array}\right. \text { at } x=2

Answers (1)

Given,

f(x)=\left\{\begin{array}{ll} 1+x, & \text { if } x \leq 2 \\ 5-x, & \text { if } x>2 \end{array}\right. \text { at } x=2

We need to check whether f(x) is continuous and differentiable at x = 2

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 = 2 if -

\lim _{h \rightarrow 0} f(2-h)=\lim _{h \rightarrow 0} f(2+h)=f(2) \\ \therefore L H L=\lim _{h \rightarrow 0} f(2-h)

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

And, f(2) = 1 + 2 = 3 {using equation 1} …(4)
From equation 2,3 and 4 we observe that:

\lim _{h \rightarrow 0} \mathrm{f}(2-\mathrm{h})=\lim _{h \rightarrow 0} \mathrm{f}(2+\mathrm{h})=\mathrm{f}(2)=3

∴ f(x) is continuous at x = 2. 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(2-h)-f(2)}{-h}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ \therefore L H D=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} \end{array}

\begin{aligned} &\Rightarrow \mathrm{LHD}=\lim _{h \rightarrow 0} \frac{1+(2-\mathrm{h})-(1+2)}{-\mathrm{h}} \quad\{\text { using equation } 1\}\\ &\Rightarrow \mathrm{LHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{3-\mathrm{h}-3}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} 1\\ &\therefore \mathrm{LHD}=1 . .0(5)\\ &\text { Now, }\\ &\lim _{\mathrm{RHD}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(2+\mathrm{h})-\mathrm{f}(2)}{\mathrm{h}}\\ &\Rightarrow \mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{5-(2+\mathrm{h})-3}{\mathrm{~h}} \quad\{\text { using equation } 1\}\\ &\Rightarrow \lim _{\mathrm{RHD}}=\underset{\mathrm{h} \rightarrow 0}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0}-1\\ &\therefore \mathrm{RHD}=-1 \ldots(6) \end{aligned}

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

(LHD at x=2) ≠ (RHD at x = 2)

∴ f(x) is not differentiable at x = 2

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