Question

Let $f\left(x\right)=\{\begin{array}{cc}cx+1;& x\le 3\\ c{x}^2-1;& x>3\end{array}$.

If $f\left(x\right)$ is continuous at $x=3$ then find the value of $c$.

Answer 1

The function $f\left(x\right)$ is given to be continuous at $x=3$.

Therefore,

$\underset{x\to 3-}{lim}f\left(x\right)=\underset{x\to 3+}{lim}f\left(x\right)$

or, $3c+1=9c-1$

or, $6c=2$

or, $c=\frac{1}{3}$.