Let $f$ be a function defined by $$f(x)=\sqrt{x^2-5x}$$ for all real number $x$ with $x\le 0$ or $x\ge 5$. Also, let $g:\Bbb R \to \Bbb R$ be a function defined by and \begin{align*} g(x)= \begin{cases} x^2+1, x>0 \\ x+1, x\le 0 \end{cases}. \end{align*} Find $f \circ g$.
Attempt: Notice that \begin{align*} (f \circ g)(x)= \begin{cases} f(x^2+1), x>0 \\ f(x+1), x\le 0 \end{cases}. \end{align*} For $x \le 0$, we get $$(f \circ g)(x) = \sqrt{x^2-3x-4}$$ whenever $x \le -1$ or $x \ge 4$. By the case, it must be happened if $x \le -1$.
For $x > 0$, we obtain $$(f \circ g)(x) = \sqrt{x^4-3x^2-4}$$ whenever $x \le -2$ or $x \ge 2$. By the case, it must be happened if $x \ge 2$.
Therefore, we conclude that \begin{align*} (f \circ g)(x)= \begin{cases} \sqrt{x^2-3x-4}, x\le -1 \\ \sqrt{x^4-3x^2-4}, x \ge 2 \end{cases}. \end{align*}
Does this correct? Thanks in advanced.