To find the inverse, we swap the variables y and x, then solve for the new y.
3a. [tex]y=\frac{3}{x-1}[/tex]
Swapping the variables: [tex]x=\frac{3}{y-1}[/tex]
Solving for y: [tex]x(y-1)=3 \\ y-1= \frac{3}{x} \\ y=1+\frac{3}{x} [/tex]
The domain of this inverse is [tex]x ≠ 0[/tex].
3b. [tex]y=x^2-1[/tex]
Swapping: [tex]x = y^2 - 1[/tex]
Solving for y: [tex]y^2 = x + 1 \\ y = \sqrt{x+1} [/tex]
The domain of this inverse is [tex]x ≥ -1[/tex].
3c. [tex]y=\sqrt[3]{\frac{x-7}{3}}[/tex]
Swapping: [tex]x=\sqrt[3]{\frac{y-7}{3}}[/tex]
Solving for y: [tex]x^3=\frac{y-7}{3} \\ y-7=3x^3 \\ y=3x^3+7[/tex]
The domain of this inverse is all real numbers.
4a. [tex]y=\frac{3}{x-1}[/tex], [tex]y=1+\frac{3}{x}[/tex]
[tex]y=\frac{3}{(1+\frac{3}{x})-1} \\ y=\frac{3}{(\frac{3}{x})} \\ y=x[/tex]
[tex]y=1+\frac{3}{(\frac{3}{x-1})} \\ y = 1+(x-1) \\ y = x[/tex]
4c. [tex]y=\sqrt[3]{\frac{x-7}{3}}[/tex], [tex]y=3x^3+7[/tex]
[tex]y=\sqrt[3]{\frac{(3x^3+7)-7}{3}} \\ y=\sqrt[3]{\frac{3x^3}{3}} \\ y=\sqrt[3]{x^3} \\ y=x[/tex]
[tex]y=3(\sqrt[3]{\frac{x-7}{3}})^3+7 \\ y = 3({\frac{x-7}{3}})+7 \\ y = (x-7)+7 \\ y=x[/tex]
