Browse Questions

# For what value of k is the following function continuous at x = 2?$f(x) = \left\{ \begin{array}{l l}2x+1, & \quad if { x \: < \: 2 } \\ k, & \quad if { x = 2 } \\ 3x-1, & \quad if { x > 2 } \end{array} \right.$

Toolbox:
• If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
$f(x) = \left\{ \begin{array}{l l}2x+1, & \quad if { x \: < \: 2 } \\ k, & \quad if { x = 2 } \\ 3x-1, & \quad if { x > 2 } \end{array} \right.$
It is continuous at $x=2$
LHL =$\lim\limits_{x\to 2^-}f(x)=\lim\limits_{x\to 2^-}(2x+1)$
Put $x=2-h$
So that $x\to 2^-\Rightarrow h\to 0$
LHL =$\lim\limits_{h\to 0}2(2h-1)+1$
$\Rightarrow 2(2-0)+1$
$\Rightarrow 4+1$
$\Rightarrow 5$
Step 2:
RHL =$\lim\limits_{x\to 2^+}f(x)=\lim\limits_{x\to 2^+}(x+1)$
Put $x=2+h$
So that when $x\to 2^+\Rightarrow h\to 0$
RHL =$\lim\limits_{h\to 0}3(2+h)-1$
$\Rightarrow 6-1$
$=5$
$f(2)=k$
If the function is continuous at $x=2$
Then LHL=RHL=f(2)
$\Rightarrow 5=5=k$
$\therefore k=5$