Answer
For $a=-3$, $g(x)$ is differentiable for all $x$-values.
Work Step by Step
$g(x)=ax$ for $x\lt0$ and $g(x)=x^2-3x$ for $x\ge0$
1) For all $x\lt0$: $$g'(x)=(ax)'=a$$
Since $a$ is a definite value, $g(x)$ is differentiable on $(-\infty,0)$.
2) For all $x\gt0$: $$g'(x)=(x^2-3x)'=2x-3$$
For all $x\gt0$, $(2x-3)$ always gives out a definite value, so $g(x)$ is differentiable on $(0,\infty)$.
3) At $x=0$:
- Left-hand derivative: $g'(0)=a$
- Right-hand derivative: $g'(0)=2\times0-3=-3$
For $g(x)$ to be differentiable at $x=0$, the left-hand derivative must be equal with the right-hand derivative. That means
$$a=-3$$
Therefore, for $a=-3$, $g(x)$ is differentiable for all $x$-values.
