Question

I need help in this!

Answer 1

For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that $a$ is such that

$\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x$

Obviously, both limits are 0, so $g$ is indeed continuous at $x=0$.

Now, for $g$ to be differentiable everywhere, its derivative $g'$ must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:

$g'(x)=\begin{cases}a&\text{for }x0\end{cases}$

and at this time, we don't know what's going on at $x=0$, so we omit that case. We want $g'$ to be continuous, so we require that

$\displaystyle\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\iff\lim_{x\to0}a=\lim_{x\to0}2x-3$

from which it follows that $a=-3$.