Find the inverse \( f^{-1} (x) \)
\(\textbf{1)}\) \( f(x)=2x-3 \)
The inverse is \( f^{-1} (x)=\displaystyle\frac{x+3}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=2x-3\)
\(\,\,\,\,\, y=2x-3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=2y-3\)
\(\,\,\,\,\, x+3=2y\)
\(\,\,\,\,\, \frac{x+3}{2}=y\)
\(\,\,\,\,\, y=\frac{x+3}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x+3}{2} \)
\(\textbf{2)}\) \( f(x)=(4x-5)^3 \)
The inverse is \( f^{-1} (x)=\displaystyle\frac{5+\sqrt[3]{x}}{4} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=(4x-5)^3\)
\(\,\,\,\,\, y=(4x-5)^3\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=(4y-5)^3\)
\(\,\,\,\,\, \sqrt[3]{x}=4y-5\)
\(\,\,\,\,\, 4y-5=\sqrt[3]{x}\)
\(\,\,\,\,\, 4y=\sqrt[3]{x}+5\)
\(\,\,\,\,\, y=\frac{5+\sqrt[3]{x}}{4}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{5+\sqrt[3]{x}}{4}\)
\(\textbf{3)}\) \( f(x)=\displaystyle\frac{x-5}{7} \)
The inverse is \( f^{-1} (x)=7x+5 \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{x-5}{7}\)
\(\,\,\,\,\, y=\frac{x-5}{7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{y-5}{7}\)
\(\,\,\,\,\, 7x=y-5\)
\(\,\,\,\,\, y=7x+5\)
\(\,\,\,\,\, f^{-1}(x)=7x+5\)
\(\textbf{4)}\) \( f(x)=\displaystyle\frac{3x+2}{x} \)
The inverse is \( f^{-1} (x)=\displaystyle\frac{2}{x-3} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\frac{3x+2}{x}\)
\(\,\,\,\,\, y=\frac{3x+2}{x}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\frac{3y+2}{y}\)
\(\,\,\,\,\, xy=3y+2\)
\(\,\,\,\,\, xy-3y=2\)
\(\,\,\,\,\, y(x-3)=2\)
\(\,\,\,\,\, y=\frac{2}{x-3}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{2}{x-3}\)
\(\textbf{5)}\) \( f(x)=\sqrt{2x+7} \)
The inverse is \( f^{-1} (x)=\displaystyle\frac{x^2-7}{2} \)
\(\text{Given}\)
\(\,\,\,\,\, f(x)=\sqrt{2x+7}\)
\(\,\,\,\,\, y=\sqrt{2x+7}\)
\(\text{Switch x and y when finding the inverse.}\)
\(\,\,\,\,\, x=\sqrt{2y+7}\)
\(\,\,\,\,\, x^2=2y+7\)
\(\,\,\,\,\, 2y=x^2-7\)
\(\,\,\,\,\, y=\frac{x^2-7}{2}\)
\(\,\,\,\,\, f^{-1}(x)=\frac{x^2-7}{2}\)
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