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  • Find which of the functions is continuous or discontinuous at the indicated points: f(x) = x^2/2

Find which of the functions is continuous or discontinuous at the indicated points:

f(x)=\left\{\begin{array}{cc} \frac{x^{2}}{2}, \text { if } & 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2 \end{array}\right.

Answers (1)

Given,

f(x)=\left\{\begin{array}{cc} \frac{x^{2}}{2}, \text { if } & 0 \leq x \leq 1 \\ 2 x^{2}-3 x+\frac{3}{2}, & \text { if } 1<x \leq 2 \end{array}\right.

We need to check its continuity at x = 1
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)

Now according to above theory-
f(x) is continuous at x = 1 if -

\begin{aligned} &\lim _{h \rightarrow 0} f(1-h)=\lim _{h \rightarrow 0} f(1+h)=f(1)\\ &\text { then, }\\ &\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \frac{(1-\mathrm{h})^{2}}{2} \quad\{\text { using equation } 1\}\\ &\therefore \mathrm{LHL}=(1-0)^{2} / 2=1 / 2 \ldots(2) \end{aligned}

Similarly, we proceed for RHL-

\begin{array}{l} \lim _{\mathrm{RHL}=} \lim _{h \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}\left(2(1+\mathrm{h})^{2}-3(1+\mathrm{h})+\frac{3}{2}\right)_{\{\text {using eqn } 1\}} \\ \Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0}\left(2\left(\mathrm{~h}^{2}+2 \mathrm{~h}+1\right)-3-3 \mathrm{~h}+\frac{3}{2}\right) \\ \Rightarrow \mathrm{RHL}=\lim _{\mathrm{h} \rightarrow 0}\left(2 \mathrm{~h}^{2}+4 \mathrm{~h}+2-3-3 \mathrm{~h}+\frac{3}{2}\right) \\ \Rightarrow \mathrm{RHL}=\lim _{h \rightarrow 0}\left(2 \mathrm{~h}^{2}+\mathrm{h}+\frac{1}{2}\right) \\ \therefore \mathrm{RHL}=2(0)^{2}+0+1 / 2=1 / 2 \ldots(3) \end{array}

And,

\\f(1)=1^{2} / 2=1 / 2\{ using \ equation \ 1\} \ldots(4)\\ \text{From the equation 2,3 and 4} \ \text{we can conclude that } \\\lim _{h \rightarrow 0} \mathrm{f}(1-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(1+\mathrm{h})=\mathrm{f}(1)=\frac{1}{2} \\\therefore f(x) \ is \ continuous \ at \ x=1

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infoexpert22

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