An algebraic function of an algebraic number yields an algebraic number, but does an algebraic function of a transcendental number always yield a transcendental number?
Yes, this is true. If $f$ is an algebraic function, then it satisfies an irreducible polynomial $p(x,f(x))=0$ with algebraic coefficients. Plugging in $x=t$, our transcendental number, we see that $p(t,f(t))=0$. But this says that a transcendental number $t$ satisfies a polyonmial with algebraic coefficients. The only way for this to be true is if that polynomial is zero: $p(x,f(t))=0$ for any $x$. But this means that $p(x,y)$ is divisible by $(y-f(t))$, contradicting irreducibility of $p$.