SOLUTION:
First, let us create a table for:
[tex]\begin{gathered} f(x)=x^2+1 \\ (-3\leq x\leq3) \end{gathered}[/tex]
[tex]\begin{gathered} f(-3)=(-3)^2+1\Rightarrow9+1=10 \\ f(-2)=(-2)^2+1\operatorname{\Rightarrow}4+1=5 \\ f(-1)=(-1)^2+1\operatorname{\Rightarrow}1+1=2 \\ f(0)=(0)^2+1\operatorname{\Rightarrow}0+1=1 \\ f(1)=(1)^2+1\operatorname{\Rightarrow}1+1=2 \\ f(2)=(2)^2+1\operatorname{\Rightarrow}4+1=5 \\ f(3)=(3)^2+1\operatorname{\Rightarrow}9+1=10 \end{gathered}[/tex]
The table of values:
The Inverse:
[tex]\begin{gathered} y=x^2+1 \\ x^2=y-1 \\ x=\sqrt{y-1} \\ f^{-1}(x)=\sqrt{x-1} \end{gathered}[/tex]
For the values with inverses:
[tex]\begin{gathered} f^{-1}(x)=\sqrt{x-1} \\ (-3\leq x\leq3) \end{gathered}[/tex][tex]\begin{gathered} f^{-1}(-3)=\sqrt{(-3)^2-1}=\sqrt{9-1}=\sqrt{8} \\ f^{-1}(-2)=\sqrt{(-2)^2-1}=\sqrt{4-1}=\sqrt{3} \\ f^{-1}(-1)=\sqrt{(-1)^2-1}=\sqrt{1-1}=\sqrt{0}=0 \\ f^{-1}(0)=\sqrt{(0)^2-1}=\sqrt{0-1}=\sqrt{-}1(discontinues) \\ f^{-1}(1)=\sqrt{(1)^2-1}=\sqrt{1-1}=\sqrt{0}=0 \\ f^{-1}(2)=\sqrt{(2)^2-1}=\sqrt{4-1}=\sqrt{3} \\ f^{-1}(3)=\sqrt{(3)^2-1}=\sqrt{9-1}=\sqrt{8} \end{gathered}[/tex]
The table of values for the inverse:
