Question

For what value of $\lambda$ is the function defined by
$f\left(x\right)=\{\begin{array}{c}\lambda (x^2-2x),\text{if}x\le 0\\ 4x+1,\text{if}x>0\end{array}$

continuous at $x=0$? What about continuity at $x=1$ ?

Answer 1

$f\left(x\right)=\{\begin{array}{c}\lambda (x^2-2x),\text{if}x\le 0\\ 4x+1,\text{if}x>0\end{array}$

If $f$ is continuous at $x=0$, then
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{+}}{lim}f\left(x\right)=f\left(0\right)$

$\Rightarrow \underset{x\to 0}{lim}\lambda (x^2-2x)=\underset{x\to 0}{lim}(4x+1)=\lambda (0^2-2x0)$

$\Rightarrow \lambda (0^2-2\cdot 0)=4\cdot 0+1=0$

$\Rightarrow 0=1=0,$ which is not possible

Therefore, there is no value of $\lambda$ for which $f$ is continuous at $x=0$

At $x=1,$

$f\left(1\right)=4x+1=4\cdot 1+1=5$

$\underset{x\to 1}{lim}(4x+1)=4\cdot 1+1=5$

$\therefore \underset{x\to 1}{lim}f\left(x\right)=f\left(1\right)$

Therefore, for any values of $\lambda$, f is continuous at $x=1$