Question
Discuss the continuity of the function $\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
✓
Answer
When x < 2, we have
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
Hence, f(x) is everywhere continuous.
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
Hence, f(x) is everywhere continuous.
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