Answer
$
f(x)=22\left(\frac{23}{11} \right)^{(x+5) / 22}$
Work Step by Step
We want to have $f(x)=a b^x$.
We can write $f(-5)=a b^{-5}=22$ and $f(17)=a b^{17}=46$. This gives
$$
\begin{aligned}
\frac{a b^{17}}{a b^{-5}} & =\frac{f(17)}{f(-5)} \\
b^{22} & =\frac{46}{22} \\
b & =\left(\frac{46}{22} \right)^{1 / 22}=\left(\frac{23}{11} \right)^{1 / 22}
\end{aligned}
$$ We determine $a$:
$$
\begin{aligned}
a b^{-5}& =22\\
a & =b^{5}\cdot 22 \\
a & =\left(\left(\frac{23}{11} \right)^{1 / 22}\right)^5\cdot 22\\
&=22\left(\frac{23}{11} \right)^{5 / 22}
\end{aligned}
$$ It follows that
$$\begin{aligned}
f(x)&=22\left(\frac{23}{11} \right)^{5 / 22}\cdot\left(\left(\frac{23}{11} \right)^{1 / 22}\right)^x\\
&=22\left(\frac{23}{11} \right)^{(x+5) / 22}
\end{aligned}$$
