Inverse functions reverse the action of a function, taking an output and returning the original input. To find an inverse algebraically, replace the function with \(y\), switch \(x\) and \(y\), and solve for \(y\). You can also verify that two functions are inverses by showing that both compositions, \(f(g(x))\) and \(g(f(x))\), simplify to \(x\).
Find the inverse.
\(\textbf{1)}\) Find the inverse of \( f(x)=2x-3 \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{x+3}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=2x-3\)
\(\,\,\,\,\, y=2x-3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=2y-3\)
\(\,\,\,\,\, x+3=2y\)
\(\,\,\,\,\, \frac{x+3}{2}=y\)
\(\,\,\,\,\, y=\frac{x+3}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x+3}{2}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=2x-3\)
\(\,\,\,\,\, y=2x-3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=2y-3\)
\(\,\,\,\,\, x+3=2y\)
\(\,\,\,\,\, \frac{x+3}{2}=y\)
\(\,\,\,\,\, y=\frac{x+3}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x+3}{2}\)
\(\textbf{2)}\) Find the inverse of \( f(x)=(4x-5)^3 \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{5+\sqrt[3]{x}}{4} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(4x-5)^3\)
\(\,\,\,\,\, y=(4x-5)^3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(4y-5)^3\)
\(\,\,\,\,\, \sqrt[3]{x}=4y-5\)
\(\,\,\,\,\, 4y=\sqrt[3]{x}+5\)
\(\,\,\,\,\, y=\frac{5+\sqrt[3]{x}}{4}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{5+\sqrt[3]{x}}{4}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(4x-5)^3\)
\(\,\,\,\,\, y=(4x-5)^3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(4y-5)^3\)
\(\,\,\,\,\, \sqrt[3]{x}=4y-5\)
\(\,\,\,\,\, 4y=\sqrt[3]{x}+5\)
\(\,\,\,\,\, y=\frac{5+\sqrt[3]{x}}{4}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{5+\sqrt[3]{x}}{4}\)
\(\textbf{3)}\) Find the inverse of \( f(x)=\displaystyle\frac{x-5}{7} \)
The inverse is \( f^{-1}(x)=7x+5 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x-5}{7}\)
\(\,\,\,\,\, y=\frac{x-5}{7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y-5}{7}\)
\(\,\,\,\,\, 7x=y-5\)
\(\,\,\,\,\, y=7x+5\)
\(\,\,\,\,\, f^{-1}(x)=7x+5\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x-5}{7}\)
\(\,\,\,\,\, y=\frac{x-5}{7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y-5}{7}\)
\(\,\,\,\,\, 7x=y-5\)
\(\,\,\,\,\, y=7x+5\)
\(\,\,\,\,\, f^{-1}(x)=7x+5\)
\(\textbf{4)}\) Find the inverse of \( f(x)=\displaystyle\frac{3x+2}{x} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{2}{x-3} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{3x+2}{x}\)
\(\,\,\,\,\, y=\frac{3x+2}{x}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{3y+2}{y}\)
\(\,\,\,\,\, xy=3y+2\)
\(\,\,\,\,\, xy-3y=2\)
\(\,\,\,\,\, y(x-3)=2\)
\(\,\,\,\,\, y=\frac{2}{x-3}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{2}{x-3}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{3x+2}{x}\)
\(\,\,\,\,\, y=\frac{3x+2}{x}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{3y+2}{y}\)
\(\,\,\,\,\, xy=3y+2\)
\(\,\,\,\,\, xy-3y=2\)
\(\,\,\,\,\, y(x-3)=2\)
\(\,\,\,\,\, y=\frac{2}{x-3}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{2}{x-3}\)
\(\textbf{5)}\) Find the inverse of \( f(x)=\sqrt{2x+7} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{x^2-7}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{2x+7}\)
\(\,\,\,\,\, y=\sqrt{2x+7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{2y+7}\)
\(\,\,\,\,\, x^2=2y+7\)
\(\,\,\,\,\, 2y=x^2-7\)
\(\,\,\,\,\, y=\frac{x^2-7}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x^2-7}{2}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{2x+7}\)
\(\,\,\,\,\, y=\sqrt{2x+7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{2y+7}\)
\(\,\,\,\,\, x^2=2y+7\)
\(\,\,\,\,\, 2y=x^2-7\)
\(\,\,\,\,\, y=\frac{x^2-7}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x^2-7}{2}\)
\(\textbf{6)}\) Find the inverse of \( f(x)=5x+8 \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{x-8}{5} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=5x+8\)
\(\,\,\,\,\, y=5x+8\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=5y+8\)
\(\,\,\,\,\, x-8=5y\)
\(\,\,\,\,\, y=\frac{x-8}{5}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x-8}{5}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=5x+8\)
\(\,\,\,\,\, y=5x+8\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=5y+8\)
\(\,\,\,\,\, x-8=5y\)
\(\,\,\,\,\, y=\frac{x-8}{5}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x-8}{5}\)
\(\textbf{7)}\) Find the inverse of \( f(x)=\displaystyle\frac{x+4}{3} \)
The inverse is \( f^{-1}(x)=3x-4 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x+4}{3}\)
\(\,\,\,\,\, y=\frac{x+4}{3}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y+4}{3}\)
\(\,\,\,\,\, 3x=y+4\)
\(\,\,\,\,\, y=3x-4\)
\(\,\,\,\,\, f^{-1}(x)=3x-4\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x+4}{3}\)
\(\,\,\,\,\, y=\frac{x+4}{3}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y+4}{3}\)
\(\,\,\,\,\, 3x=y+4\)
\(\,\,\,\,\, y=3x-4\)
\(\,\,\,\,\, f^{-1}(x)=3x-4\)
\(\textbf{8)}\) Find the inverse of \( f(x)=(2x+1)^3 \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{\sqrt[3]{x}-1}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(2x+1)^3\)
\(\,\,\,\,\, y=(2x+1)^3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(2y+1)^3\)
\(\,\,\,\,\, \sqrt[3]{x}=2y+1\)
\(\,\,\,\,\, \sqrt[3]{x}-1=2y\)
\(\,\,\,\,\, y=\frac{\sqrt[3]{x}-1}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{\sqrt[3]{x}-1}{2}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(2x+1)^3\)
\(\,\,\,\,\, y=(2x+1)^3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(2y+1)^3\)
\(\,\,\,\,\, \sqrt[3]{x}=2y+1\)
\(\,\,\,\,\, \sqrt[3]{x}-1=2y\)
\(\,\,\,\,\, y=\frac{\sqrt[3]{x}-1}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{\sqrt[3]{x}-1}{2}\)
\(\textbf{9)}\) Find the inverse of \( f(x)=\sqrt[3]{x-6} \)
The inverse is \( f^{-1}(x)=x^3+6 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt[3]{x-6}\)
\(\,\,\,\,\, y=\sqrt[3]{x-6}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt[3]{y-6}\)
\(\,\,\,\,\, x^3=y-6\)
\(\,\,\,\,\, y=x^3+6\)
\(\,\,\,\,\, f^{-1}(x)=x^3+6\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt[3]{x-6}\)
\(\,\,\,\,\, y=\sqrt[3]{x-6}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt[3]{y-6}\)
\(\,\,\,\,\, x^3=y-6\)
\(\,\,\,\,\, y=x^3+6\)
\(\,\,\,\,\, f^{-1}(x)=x^3+6\)
\(\textbf{10)}\) Find the inverse of \( f(x)=\sqrt{3x-2} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{x^2+2}{3} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{3x-2}\)
\(\,\,\,\,\, y=\sqrt{3x-2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{3y-2}\)
\(\,\,\,\,\, x^2=3y-2\)
\(\,\,\,\,\, x^2+2=3y\)
\(\,\,\,\,\, y=\frac{x^2+2}{3}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x^2+2}{3}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{3x-2}\)
\(\,\,\,\,\, y=\sqrt{3x-2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{3y-2}\)
\(\,\,\,\,\, x^2=3y-2\)
\(\,\,\,\,\, x^2+2=3y\)
\(\,\,\,\,\, y=\frac{x^2+2}{3}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x^2+2}{3}\)
\(\textbf{11)}\) Find the inverse of \( f(x)=x^2+5,\;x\ge0 \)
The inverse is \( f^{-1}(x)=\sqrt{x-5} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=x^2+5,\;x\ge0\)
\(\,\,\,\,\, y=x^2+5\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=y^2+5\)
\(\,\,\,\,\, x-5=y^2\)
\(\,\,\,\,\, y=\sqrt{x-5}\)
\(\,\,\,\,\, f^{-1}(x)=\sqrt{x-5}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=x^2+5,\;x\ge0\)
\(\,\,\,\,\, y=x^2+5\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=y^2+5\)
\(\,\,\,\,\, x-5=y^2\)
\(\,\,\,\,\, y=\sqrt{x-5}\)
\(\,\,\,\,\, f^{-1}(x)=\sqrt{x-5}\)
\(\textbf{12)}\) Find the inverse of \( f(x)=(x-2)^3+1 \)
The inverse is \( f^{-1}(x)=2+\sqrt[3]{x-1} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(x-2)^3+1\)
\(\,\,\,\,\, y=(x-2)^3+1\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(y-2)^3+1\)
\(\,\,\,\,\, x-1=(y-2)^3\)
\(\,\,\,\,\, \sqrt[3]{x-1}=y-2\)
\(\,\,\,\,\, y=2+\sqrt[3]{x-1}\)
\(\,\,\,\,\, f^{-1}(x)=2+\sqrt[3]{x-1}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(x-2)^3+1\)
\(\,\,\,\,\, y=(x-2)^3+1\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(y-2)^3+1\)
\(\,\,\,\,\, x-1=(y-2)^3\)
\(\,\,\,\,\, \sqrt[3]{x-1}=y-2\)
\(\,\,\,\,\, y=2+\sqrt[3]{x-1}\)
\(\,\,\,\,\, f^{-1}(x)=2+\sqrt[3]{x-1}\)
\(\textbf{13)}\) Find the inverse of \( f(x)=\displaystyle\frac{1}{x+2} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{1}{x}-2 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{1}{x+2}\)
\(\,\,\,\,\, y=\frac{1}{x+2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{1}{y+2}\)
\(\,\,\,\,\, x(y+2)=1\)
\(\,\,\,\,\, xy+2x=1\)
\(\,\,\,\,\, xy=1-2x\)
\(\,\,\,\,\, y=\frac{1-2x}{x}\)
\(\,\,\,\,\, y=\frac{1}{x}-2\)
\(\,\,\,\,\, f^{-1}(x)=\frac{1}{x}-2\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{1}{x+2}\)
\(\,\,\,\,\, y=\frac{1}{x+2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{1}{y+2}\)
\(\,\,\,\,\, x(y+2)=1\)
\(\,\,\,\,\, xy+2x=1\)
\(\,\,\,\,\, xy=1-2x\)
\(\,\,\,\,\, y=\frac{1-2x}{x}\)
\(\,\,\,\,\, y=\frac{1}{x}-2\)
\(\,\,\,\,\, f^{-1}(x)=\frac{1}{x}-2\)
\(\textbf{14)}\) Find the inverse of \( f(x)=\displaystyle\frac{3x-4}{x+2} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{2x+4}{3-x} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{3x-4}{x+2}\)
\(\,\,\,\,\, y=\frac{3x-4}{x+2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{3y-4}{y+2}\)
\(\,\,\,\,\, x(y+2)=3y-4\)
\(\,\,\,\,\, xy+2x=3y-4\)
\(\,\,\,\,\, xy-3y=-4-2x\)
\(\,\,\,\,\, y(x-3)=-4-2x\)
\(\,\,\,\,\, y=\frac{-4-2x}{x-3}\)
\(\,\,\,\,\, y=\frac{2x+4}{3-x}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{2x+4}{3-x}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{3x-4}{x+2}\)
\(\,\,\,\,\, y=\frac{3x-4}{x+2}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{3y-4}{y+2}\)
\(\,\,\,\,\, x(y+2)=3y-4\)
\(\,\,\,\,\, xy+2x=3y-4\)
\(\,\,\,\,\, xy-3y=-4-2x\)
\(\,\,\,\,\, y(x-3)=-4-2x\)
\(\,\,\,\,\, y=\frac{-4-2x}{x-3}\)
\(\,\,\,\,\, y=\frac{2x+4}{3-x}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{2x+4}{3-x}\)
\(\textbf{15)}\) Find the inverse of \( f(x)=\displaystyle\frac{x+1}{2x-3} \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{3x+1}{2x-1} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x+1}{2x-3}\)
\(\,\,\,\,\, y=\frac{x+1}{2x-3}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y+1}{2y-3}\)
\(\,\,\,\,\, x(2y-3)=y+1\)
\(\,\,\,\,\, 2xy-3x=y+1\)
\(\,\,\,\,\, 2xy-y=3x+1\)
\(\,\,\,\,\, y(2x-1)=3x+1\)
\(\,\,\,\,\, y=\frac{3x+1}{2x-1}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{3x+1}{2x-1}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x+1}{2x-3}\)
\(\,\,\,\,\, y=\frac{x+1}{2x-3}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y+1}{2y-3}\)
\(\,\,\,\,\, x(2y-3)=y+1\)
\(\,\,\,\,\, 2xy-3x=y+1\)
\(\,\,\,\,\, 2xy-y=3x+1\)
\(\,\,\,\,\, y(2x-1)=3x+1\)
\(\,\,\,\,\, y=\frac{3x+1}{2x-1}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{3x+1}{2x-1}\)
\(\textbf{16)}\) Find the inverse of \( f(x)=4-2x \)
The inverse is \( f^{-1}(x)=\displaystyle\frac{4-x}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=4-2x\)
\(\,\,\,\,\, y=4-2x\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=4-2y\)
\(\,\,\,\,\, x-4=-2y\)
\(\,\,\,\,\, y=\frac{4-x}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{4-x}{2}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=4-2x\)
\(\,\,\,\,\, y=4-2x\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=4-2y\)
\(\,\,\,\,\, x-4=-2y\)
\(\,\,\,\,\, y=\frac{4-x}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{4-x}{2}\)
\(\textbf{17)}\) Find the inverse of \( f(x)=\displaystyle\frac{7}{x-1}+3 \)
The inverse is \( f^{-1}(x)=1+\displaystyle\frac{7}{x-3} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{7}{x-1}+3\)
\(\,\,\,\,\, y=\frac{7}{x-1}+3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{7}{y-1}+3\)
\(\,\,\,\,\, x-3=\frac{7}{y-1}\)
\(\,\,\,\,\, (x-3)(y-1)=7\)
\(\,\,\,\,\, y-1=\frac{7}{x-3}\)
\(\,\,\,\,\, y=1+\frac{7}{x-3}\)
\(\,\,\,\,\, f^{-1}(x)=1+\frac{7}{x-3}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{7}{x-1}+3\)
\(\,\,\,\,\, y=\frac{7}{x-1}+3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{7}{y-1}+3\)
\(\,\,\,\,\, x-3=\frac{7}{y-1}\)
\(\,\,\,\,\, (x-3)(y-1)=7\)
\(\,\,\,\,\, y-1=\frac{7}{x-3}\)
\(\,\,\,\,\, y=1+\frac{7}{x-3}\)
\(\,\,\,\,\, f^{-1}(x)=1+\frac{7}{x-3}\)
\(\textbf{18)}\) Find the inverse of \( f(x)=\sqrt{x+5}-2 \)
The inverse is \( f^{-1}(x)=(x+2)^2-5 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{x+5}-2\)
\(\,\,\,\,\, y=\sqrt{x+5}-2\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{y+5}-2\)
\(\,\,\,\,\, x+2=\sqrt{y+5}\)
\(\,\,\,\,\, (x+2)^2=y+5\)
\(\,\,\,\,\, y=(x+2)^2-5\)
\(\,\,\,\,\, f^{-1}(x)=(x+2)^2-5\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{x+5}-2\)
\(\,\,\,\,\, y=\sqrt{x+5}-2\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{y+5}-2\)
\(\,\,\,\,\, x+2=\sqrt{y+5}\)
\(\,\,\,\,\, (x+2)^2=y+5\)
\(\,\,\,\,\, y=(x+2)^2-5\)
\(\,\,\,\,\, f^{-1}(x)=(x+2)^2-5\)
\(\textbf{19)}\) Find the inverse of \( f(x)=(x-3)^2,\;x\ge3 \)
The inverse is \( f^{-1}(x)=3+\sqrt{x} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(x-3)^2,\;x\ge3\)
\(\,\,\,\,\, y=(x-3)^2\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(y-3)^2\)
\(\,\,\,\,\, \sqrt{x}=y-3\)
\(\,\,\,\,\, y=3+\sqrt{x}\)
\(\,\,\,\,\, f^{-1}(x)=3+\sqrt{x}\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(x-3)^2,\;x\ge3\)
\(\,\,\,\,\, y=(x-3)^2\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(y-3)^2\)
\(\,\,\,\,\, \sqrt{x}=y-3\)
\(\,\,\,\,\, y=3+\sqrt{x}\)
\(\,\,\,\,\, f^{-1}(x)=3+\sqrt{x}\)
\(\textbf{20)}\) Find the inverse of \( f(x)=2\sqrt[3]{x+1}-5 \)
The inverse is \( f^{-1}(x)=\left(\displaystyle\frac{x+5}{2}\right)^3-1 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=2\sqrt[3]{x+1}-5\)
\(\,\,\,\,\, y=2\sqrt[3]{x+1}-5\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=2\sqrt[3]{y+1}-5\)
\(\,\,\,\,\, x+5=2\sqrt[3]{y+1}\)
\(\,\,\,\,\, \frac{x+5}{2}=\sqrt[3]{y+1}\)
\(\,\,\,\,\, \left(\frac{x+5}{2}\right)^3=y+1\)
\(\,\,\,\,\, y=\left(\frac{x+5}{2}\right)^3-1\)
\(\,\,\,\,\, f^{-1}(x)=\left(\frac{x+5}{2}\right)^3-1\)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=2\sqrt[3]{x+1}-5\)
\(\,\,\,\,\, y=2\sqrt[3]{x+1}-5\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=2\sqrt[3]{y+1}-5\)
\(\,\,\,\,\, x+5=2\sqrt[3]{y+1}\)
\(\,\,\,\,\, \frac{x+5}{2}=\sqrt[3]{y+1}\)
\(\,\,\,\,\, \left(\frac{x+5}{2}\right)^3=y+1\)
\(\,\,\,\,\, y=\left(\frac{x+5}{2}\right)^3-1\)
\(\,\,\,\,\, f^{-1}(x)=\left(\frac{x+5}{2}\right)^3-1\)
Verify that the functions are inverses using composition.
\(\textbf{21)}\) Verify that \(f(x)=2x+3\) and \(g(x)=\displaystyle\frac{x-3}{2}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{x-3}{2}\right)\)
\(=2\left(\frac{x-3}{2}\right)+3\)
\(=x-3+3\)
\(=x\)
\(g(f(x))\)
\(g(2x+3)\)
\(=\frac{(2x+3)-3}{2}\)
\(=\frac{2x}{2}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{x-3}{2}\right)\)
\(=2\left(\frac{x-3}{2}\right)+3\)
\(=x-3+3\)
\(=x\)
\(g(f(x))\)
\(g(2x+3)\)
\(=\frac{(2x+3)-3}{2}\)
\(=\frac{2x}{2}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{22)}\) Verify that \(f(x)=5x-4\) and \(g(x)=\displaystyle\frac{x+4}{5}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{x+4}{5}\right)\)
\(=5\left(\frac{x+4}{5}\right)-4\)
\(=x+4-4\)
\(=x\)
\(g(f(x))\)
\(g(5x-4)\)
\(=\frac{(5x-4)+4}{5}\)
\(=\frac{5x}{5}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{x+4}{5}\right)\)
\(=5\left(\frac{x+4}{5}\right)-4\)
\(=x+4-4\)
\(=x\)
\(g(f(x))\)
\(g(5x-4)\)
\(=\frac{(5x-4)+4}{5}\)
\(=\frac{5x}{5}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{23)}\) Verify that \(f(x)=\displaystyle\frac{x+2}{3}\) and \(g(x)=3x-2\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f(3x-2)\)
\(=\frac{(3x-2)+2}{3}\)
\(=\frac{3x}{3}\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{x+2}{3}\right)\)
\(=3\left(\frac{x+2}{3}\right)-2\)
\(=x+2-2\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f(3x-2)\)
\(=\frac{(3x-2)+2}{3}\)
\(=\frac{3x}{3}\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{x+2}{3}\right)\)
\(=3\left(\frac{x+2}{3}\right)-2\)
\(=x+2-2\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{24)}\) Verify that \(f(x)=(x-1)^3\) and \(g(x)=\sqrt[3]{x}+1\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f(\sqrt[3]{x}+1)\)
\(=\left((\sqrt[3]{x}+1)-1\right)^3\)
\(=(\sqrt[3]{x})^3\)
\(=x\)
\(g(f(x))\)
\(g((x-1)^3)\)
\(=\sqrt[3]{(x-1)^3}+1\)
\(=x-1+1\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f(\sqrt[3]{x}+1)\)
\(=\left((\sqrt[3]{x}+1)-1\right)^3\)
\(=(\sqrt[3]{x})^3\)
\(=x\)
\(g(f(x))\)
\(g((x-1)^3)\)
\(=\sqrt[3]{(x-1)^3}+1\)
\(=x-1+1\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{25)}\) Verify that \(f(x)=\sqrt[3]{2x+5}\) and \(g(x)=\displaystyle\frac{x^3-5}{2}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{x^3-5}{2}\right)\)
\(=\sqrt[3]{2\left(\frac{x^3-5}{2}\right)+5}\)
\(=\sqrt[3]{x^3-5+5}\)
\(=\sqrt[3]{x^3}\)
\(=x\)
\(g(f(x))\)
\(g(\sqrt[3]{2x+5})\)
\(=\frac{(\sqrt[3]{2x+5})^3-5}{2}\)
\(=\frac{2x+5-5}{2}\)
\(=\frac{2x}{2}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{x^3-5}{2}\right)\)
\(=\sqrt[3]{2\left(\frac{x^3-5}{2}\right)+5}\)
\(=\sqrt[3]{x^3-5+5}\)
\(=\sqrt[3]{x^3}\)
\(=x\)
\(g(f(x))\)
\(g(\sqrt[3]{2x+5})\)
\(=\frac{(\sqrt[3]{2x+5})^3-5}{2}\)
\(=\frac{2x+5-5}{2}\)
\(=\frac{2x}{2}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{26)}\) Verify that \(f(x)=\displaystyle\frac{1}{x}+4\) and \(g(x)=\displaystyle\frac{1}{x-4}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{1}{x-4}\right)\)
\(=\frac{1}{\frac{1}{x-4}}+4\)
\(=x-4+4\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{1}{x}+4\right)\)
\(=\frac{1}{\left(\frac{1}{x}+4\right)-4}\)
\(=\frac{1}{\frac{1}{x}}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{1}{x-4}\right)\)
\(=\frac{1}{\frac{1}{x-4}}+4\)
\(=x-4+4\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{1}{x}+4\right)\)
\(=\frac{1}{\left(\frac{1}{x}+4\right)-4}\)
\(=\frac{1}{\frac{1}{x}}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{27)}\) Verify that \(f(x)=\displaystyle\frac{2x+1}{x-3}\) and \(g(x)=\displaystyle\frac{3x+1}{x-2}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{3x+1}{x-2}\right)\)
\(=\frac{2\left(\frac{3x+1}{x-2}\right)+1}{\left(\frac{3x+1}{x-2}\right)-3}\)
\(=\frac{\frac{6x+2+x-2}{x-2}}{\frac{3x+1-3x+6}{x-2}}\)
\(=\frac{\frac{7x}{x-2}}{\frac{7}{x-2}}\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{2x+1}{x-3}\right)\)
\(=\frac{3\left(\frac{2x+1}{x-3}\right)+1}{\left(\frac{2x+1}{x-3}\right)-2}\)
\(=\frac{\frac{6x+3+x-3}{x-3}}{\frac{2x+1-2x+6}{x-3}}\)
\(=\frac{\frac{7x}{x-3}}{\frac{7}{x-3}}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{3x+1}{x-2}\right)\)
\(=\frac{2\left(\frac{3x+1}{x-2}\right)+1}{\left(\frac{3x+1}{x-2}\right)-3}\)
\(=\frac{\frac{6x+2+x-2}{x-2}}{\frac{3x+1-3x+6}{x-2}}\)
\(=\frac{\frac{7x}{x-2}}{\frac{7}{x-2}}\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{2x+1}{x-3}\right)\)
\(=\frac{3\left(\frac{2x+1}{x-3}\right)+1}{\left(\frac{2x+1}{x-3}\right)-2}\)
\(=\frac{\frac{6x+3+x-3}{x-3}}{\frac{2x+1-2x+6}{x-3}}\)
\(=\frac{\frac{7x}{x-3}}{\frac{7}{x-3}}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{28)}\) Verify that \(f(x)=7-3x\) and \(g(x)=\displaystyle\frac{7-x}{3}\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{7-x}{3}\right)\)
\(=7-3\left(\frac{7-x}{3}\right)\)
\(=7-(7-x)\)
\(=7-7+x\)
\(=x\)
\(g(f(x))\)
\(g(7-3x)\)
\(=\frac{7-(7-3x)}{3}\)
\(=\frac{3x}{3}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{7-x}{3}\right)\)
\(=7-3\left(\frac{7-x}{3}\right)\)
\(=7-(7-x)\)
\(=7-7+x\)
\(=x\)
\(g(f(x))\)
\(g(7-3x)\)
\(=\frac{7-(7-3x)}{3}\)
\(=\frac{3x}{3}\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{29)}\) Verify that \(f(x)=(x+4)^3-2\) and \(g(x)=\sqrt[3]{x+2}-4\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f(\sqrt[3]{x+2}-4)\)
\(=\left((\sqrt[3]{x+2}-4)+4\right)^3-2\)
\(=(\sqrt[3]{x+2})^3-2\)
\(=x+2-2\)
\(=x\)
\(g(f(x))\)
\(g((x+4)^3-2)\)
\(=\sqrt[3]{((x+4)^3-2)+2}-4\)
\(=\sqrt[3]{(x+4)^3}-4\)
\(=x+4-4\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f(\sqrt[3]{x+2}-4)\)
\(=\left((\sqrt[3]{x+2}-4)+4\right)^3-2\)
\(=(\sqrt[3]{x+2})^3-2\)
\(=x+2-2\)
\(=x\)
\(g(f(x))\)
\(g((x+4)^3-2)\)
\(=\sqrt[3]{((x+4)^3-2)+2}-4\)
\(=\sqrt[3]{(x+4)^3}-4\)
\(=x+4-4\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(\textbf{30)}\) Verify that \(f(x)=\displaystyle\frac{4}{x+1}-2\) and \(g(x)=\displaystyle\frac{4}{x+2}-1\) are inverses using composition.
Yes. \(f(g(x))=x\) and \(g(f(x))=x\), so the functions are inverses.
\(f(g(x))\)
\(f\left(\frac{4}{x+2}-1\right)\)
\(=\frac{4}{\left(\frac{4}{x+2}-1\right)+1}-2\)
\(=\frac{4}{\frac{4}{x+2}}-2\)
\(=x+2-2\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{4}{x+1}-2\right)\)
\(=\frac{4}{\left(\frac{4}{x+1}-2\right)+2}-1\)
\(=\frac{4}{\frac{4}{x+1}}-1\)
\(=x+1-1\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
\(f(g(x))\)
\(f\left(\frac{4}{x+2}-1\right)\)
\(=\frac{4}{\left(\frac{4}{x+2}-1\right)+1}-2\)
\(=\frac{4}{\frac{4}{x+2}}-2\)
\(=x+2-2\)
\(=x\)
\(g(f(x))\)
\(g\left(\frac{4}{x+1}-2\right)\)
\(=\frac{4}{\left(\frac{4}{x+1}-2\right)+2}-1\)
\(=\frac{4}{\frac{4}{x+1}}-1\)
\(=x+1-1\)
\(=x\)
\(\therefore\ f(g(x))=x\text{ and }g(f(x))=x\)
\(\text{The functions are inverses.}\)
See Related Pages\(\)
\(\bullet\text{ Composite Functions}\)
\(\,\,\,\,\,\,\,\,(f\circ g)(x)=f(g(x))\)
\(\bullet\text{ Inverse Functions and Relations}\)
\(\,\,\,\,\,\,\,\,f^{-1}(x)=\frac{x+3}{2}\)
\(\bullet\text{ Operations of Functions}\)
\(\,\,\,\,\,\,\,\,(f+g)(3)=f(3)+g(3)\)