Let $f(x)=\begin{cases}2x+a,&x\geqslant -1\\bx^2+3, &x\lt -1\end{cases} \;\\g(x)=\begin{cases}x+4, &0\leqslant x\leqslant 4\\3x-2,& -2\lt x\lt 0\end{cases}$
Find the range of $a$ and $b$ so that $g(f(x))$ is not defined.
My attempt:
By the domain of $g(x)$, we can conclude that $f(x)$ must be $(-2,4].$ So if the least value of $f(x)$ is greater than $4$, then $g(x)$ won't be defined. So at $x=-1$ $f(x)=-2+a.$ So, $-2+a$ must be greater than $4$ which gives $a$ must be greater than $6$. For $b\gt 1$ it would always be greater than $4$. But why am I wrong? Answer is given as $a\in(10,\infty)$ and $b\in(5,\infty)$