Answer
(a) $(-1,\infty)$.
(b) $ 0$, $(7,0)$.
(c) $ 3$, $(3,-1)$.
(d) $ 7$.
Work Step by Step
Given $F(x)=log_2(x+1)-3$, we have:
(a) the domain $x+1\gt0$ or $(-1,\infty)$.
(b) $F(7)=log_2(7+1)-3=0$ giving point $(7,0)$.
(c) $F(x)=log_2(x+1)-3=-1 \Longrightarrow x+1=2^2 \Longrightarrow x=3$ giving point $(3,-1)$.
(d) $F(x)=log_2(x+1)-3=0 \Longrightarrow x+1=2^3 \Longrightarrow x=7$.
