For what value of is the function f(x)= (x^2-2x),&if x 0 4x+1,&if… - Vidyadip

Question

For what value of $\lambda$ is the function $\text{f(x)}=\begin{cases}\lambda(\text{x}^2-2\text{x}),&\text{if }\text{ x}\leq0\\4\text{x}+1,&\text{if }\text{ x}>0\end{cases}$ continuous at x = 0? What about continuity at $\text{x}=\pm1?$

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Answer

The given function f is $\text{f(x)}=\begin{cases}\lambda(\text{x}^2-2\text{x}),&\text{if }\text{ x}\leq0\\4\text{x}+1,&\text{if }\text{ x}>0\end{cases}$
If f is continuous at x = 0, then
$\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$
$\Rightarrow\lim_\limits{\text{x}\rightarrow0^-}\lambda(\text{x}^2-2\text{x})=\lim_\limits{\text{x}\rightarrow0^+}(4\text{x}+1)=\lambda(0^2-2\times0)$
$\Rightarrow\lambda(0^2-2\times0)=4\times0+1=0$
$\Rightarrow0=1=0$ which is not possible
Therefore, there is no value of $\lambda$ for which f(x) is continuous at x = 0
At x = 1
$\Rightarrow\text{f}(1)=4\text{x}+1=4\times1+1=5$
$\Rightarrow\lim_\limits{\text{x}\rightarrow1}(4\text{x}+1)=4\times1+1=5$
$\therefore\ \lim_\limits{\text{x}\rightarrow1}\text{f(x)}=\text{f}(1)$
Therefore, for any value of $\lambda,$ f is continuous at x = 1
At x = -1 we have,
$\Rightarrow\text{f}(-1)=\lambda(1+2)=3\lambda$
$\Rightarrow\lim_\limits{\text{x}\rightarrow-1}\lambda(1+2)=3\lambda$
$\therefore\ \lim_\limits{\text{x}\rightarrow-1}\text{f(x)}=\text{f}(-1)$
Therefore, for any values of $\lambda,$ f is continuous at x = -1

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