Answer
$$
g(x)=\left\{\begin{array}{ll}{\frac{2 x^{2}-x-15}{x-3}} & {\text { if } x \neq 3} \\ {k x-1} & {\text { if } x=3}\end{array}\right.
$$
The value of the constant $k$ that makes the function continuous is .
$k=4 $
Work Step by Step
$$
g(x)=\left\{\begin{array}{ll}{\frac{2 x^{2}-x-15}{x-3}} & {\text { if } x \neq 3} \\ {k x-1} & {\text { if } x=3}\end{array}\right.
$$
This rational function is discontinuous wherever the denominator is zero.
There is a discontinuity when $x=3$ .
We investigate at $x=3$ first.
$$
\begin{aligned}
\lim _{x \rightarrow 3} g(x) &=\lim _{x \rightarrow 3}\frac{2 x^{2}-x-15}{x-3} \\
&=\lim _{x \rightarrow 3}\frac{(2 x+5)(x-3)}{x-3} \\
&=\lim _{x \rightarrow 3}(2 x+5)\\
&=2 (3)+5 \\
&=11
\end{aligned}
$$
Furthermore,
$$
g(3)= k(3)-1=3k-1
$$
since, $g$ is continuous, then we have
$$
g(3)=\lim _{x \rightarrow 3} g(x)
$$
so we obtain
$$
\begin{aligned} 3k -1&=11 \\ 3k &=12 \\4 &=k \end{aligned}
$$
Thus the value of the constant $k$ that makes the function continuous is .
$k=4$
