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 }x<0\\2x-3&\text{for }x>0\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`.