Find the values of k so that the function f is continuous at the indicated point.

The given function $f$ is $f(x)=\left\{\begin{array}{l}k x+1, \text { if } x \leq 5 \\ 3 x-5, \text { if } x>5\end{array}\right.$

The given function f is continuous at x = 5, if f is defined at x = 5 and if the value of f at x = 5 equals the limit of f at x = 5

It is evident that $f$ is defined at $x=5$ and $f(5)=k x+1=5 k+1$

$\lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5)$

$\Rightarrow \lim _{x \rightarrow 5}(k x+1)=\lim _{x \rightarrow 5^{+}}(3 x-5)=5 k+1$

$\Rightarrow 5 k+1=15-5=5 k+1$

$\Rightarrow 5 k+1=10$

$\Rightarrow 5 k=9$

$\Rightarrow k=\frac{9}{5}$

Therefore, the required value of $k$ is $\frac{9}{5}$.