Answer
$$
g(x)=\left\{\begin{array}{ll}{ x^{3}+k} & {\text { if } x \leq 3} \\ {kx-5} & {\text { if } x>3}\end{array}\right.
$$
The value of the constant $k$ that makes the function continuous is .
$k=16 $
Work Step by Step
$$
g(x)=\left\{\begin{array}{ll}{ x^{3}+k} & {\text { if } x \leq 3} \\ {kx-5} & {\text { if } x>3}\end{array}\right.
$$
Since each piece of this function is a polynomial, the only $x$-value where
$g$ might be discontinuous here is 3. We investigate at $x=3$ first. From the left, where $x$-values are less than $3$,
$$
\lim _{x \rightarrow 3^{-}} g(x)=\lim _{x \rightarrow 3^{-}}\left(x^{3}+k\right)=(3)^{3}+k=27+k
$$
From the right, where $x$-values are greater than 3,
$$
\lim _{x \rightarrow 3^{+}} g(x)=\lim _{x \rightarrow 3^{+}}\left(kx-5\right)=3k-5
$$
Furthermore,
$$
g(3)= (3)^{3}+k=27+k
$$
since, $g$ is continuous, then we have
$$
g(3)=\lim _{x \rightarrow 3^{+}} g(x)=\lim _{x \rightarrow 3^{-}}g(x)
$$
so that
$$
x^{3}+k=kx-5 \text { for } x=3
$$
we obtain
$$
\begin{aligned} (3)^{3}+k &=(3)k-5 \\ 27+k &=3k-5 \\ 3 2 &=2k \\ k &=16 \end{aligned}
$$
Thus the value of the constant $k$ that makes the function continuous is .
$k=16 $
