Examine the continuity of the function f(x)= 3x-2 &, if x 0 +1 &,… - Vidyadip

Question

Examine the continuity of the function
$\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0$
Also sketch the graph of this function.

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Answer

Given function is, $\text{f}\text{(x)}=\begin{cases}3\text{x}-2 &, \text{ if x} \leq 0\\\text{x}+1 &, \text{ if x} > 0\end{cases}\text{at x}=0\ ...(\text{ii})$ We need to check whether f(x) is continuous at x = 0 or not. For this we need to check L.H.L, R.H.L and value of function at x = 0 Clearly, f(0) = 3*0 - 2 = -2 [from equation ii] $\text{L.H.L}=\lim\limits_{\text{h} \rightarrow 0}(0-\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(-\text{h})$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{3(-\text{h})-2\big\}=-2$ $\text{R.H.L}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0+\text{h})=\lim\limits_{\text{h} \rightarrow 0}\text{f(h)}$ $=\lim\limits_{\text{h} \rightarrow 0}\big\{\text{h}+1\big\}=0+1=1$ As, $\text{L.H.L}\neq\text{R.H.L}$ $\therefore\ \text{f(x)}$ is discontinuous at x = 0 This can also be proved by plotting f(x) on cartesian plane. For x >0 ,we need to plot y = x + 1 put y = 0, we get x = -1 and for second point we put x = 0 and thus get y = 1 Two points are enough to plot the straight line. Two coordinates are (-1, 0) and (0, 1) For $\text{x}\leq0,$ we need to plot y = 3x - 2 put x = 0 then y = -2 On putting y = 0 we get $\text{x}=\frac{2}{3}$ Two coordinates are (0, -2) and $\Big(\frac{2}{3},0\Big)$ It can be seen from graph that there is breakage in curve at (0, 0) Thus, it is discontinuous at x = 0

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