The shortest path touching a line

1. Derivative.
$$L'(x)=\frac x{\sqrt{x^2+1}}-\frac{4-x}{\sqrt{(4-x)^2+9}}.$$
$L'(x)=0$ means $\dfrac x{\sqrt{x^2+1}}=\dfrac{4-x}{\sqrt{(4-x)^2+9}}$. Both sides must have the same sign, so $x$ and $4-x$ have the same sign, i.e. $x\in(0,4)$. Squaring,
$$x^2\bigl((4-x)^2+9\bigr)=(4-x)^2(x^2+1)\ \Longrightarrow\ 9x^2=(4-x)^2\ \Longrightarrow\ 3x=\pm(4-x),$$
and the condition $x\in(0,4)$ rejects $3x=-(4-x)$ (i.e. $x=-2$) and keeps $3x=4-x$, i.e. $x=1$.
$L$ is convex, as a sum of two functions of the form $x\mapsto\sqrt{(x-c)^2+d^2}$ (the norm of an affine function of $x$), so this critical point is the global minimum, and
$$L(1)=\sqrt2+\sqrt{18}=4\sqrt2.$$

2. Reflection. Let $B'=(4,-3)$. For every $M$ on the axis, $MB=MB'$, so $AM+MB=AM+MB'\ge AB'=\sqrt{16+16}=4\sqrt2$, with equality when $M$ is on the segment $AB'$. This segment lies on the line $y=1-x$, which meets the axis at $x=1$, in agreement with part 1.

Remark. At the optimum, the angles of incidence and reflection are equal: this is the law of reflection of light (Heron of Alexandria's principle).