Show algebraically that the functions

f

and

g

shown below are inverse functions.

f ( x ) = \sqrt [ 3 ] { 8 x - 7 } , \quad g ( x ) = \frac { x ^ { 3 } + 7 } { 8 }

A)

\begin{aligned}f(g(x)) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right)-7} & g(f(x)) &=\frac{(\sqrt[3]{8 x-7})^{3}+7}{8} \\&=\sqrt[3]{\left(x^{3}+56\right)-56} & & \frac{8 x-7^{3}+7^{3}}{8} \\&=\sqrt[3]{x^{3}+56-56} & &=\frac{8 x}{8} \\&=\sqrt[3]{x^{3}} & &\\& =x&=x\\\end{aligned}

B)

\begin{aligned}f(g(x)) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right)-7} & g(f(x)) &=\frac{(\sqrt[3]{8 x-7})^{3}+7}{8} \\&=\sqrt[3]{\left(x^{3}+7\right)-7} & &=\frac{8 x-7+7}{8} \\&=\sqrt[3]{x^{3}+7-7} & &=\frac{8 x}{8} \\&=\sqrt[3]{x^{3}} & \\&=x&=x\\\end{aligned}

C)

\begin{array}{l}\begin{array}{lll}f(g(x)) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right)-7} \\&=\sqrt[3]{\left(\frac{8 x^{3}+7}{8}\right)-7} \\&=\sqrt[3]{x^{3}+7-7} \\&=\sqrt[3]{x^{3}} \\&=x\end{array}\quad \quad \begin{array}{lll}g(f(x)) &=\frac{(\sqrt[3]{8 x-7})^{3}+7}{8} \\&=\frac{8^{3} x-7+7}{8^{3}} \\&=\frac{8^{3} x}{8^{3}} \\\\&=x\end{array}\end{array}

D)

\begin{aligned}f(g(x)) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right)-7} & g(f(x)) &=\frac{(\sqrt[3]{8 x-7})^{3}+7}{8} \\&=\sqrt[3]{\left(8 x^{3}+56\right)-56} & &=\frac{8^{3} x-7^{3}+7^{3}}{8^{3}} \\&=\sqrt[3]{8 x^{3}+56-56} & & =\frac{8^{3} x}{8^{3}} \\&=\sqrt[3]{8 x^{3}} & & \\&=x & &=x \end{aligned}

E)

\begin{aligned}f(g(x)) &=\sqrt[3]{8\left(\frac{x^{3}+7}{8}\right)-7} & g(f(x)) &=\frac{(\sqrt[3]{8 x-7})^{3}+7}{8} \\&=\sqrt[3]{\left(x^{3}+\frac{7}{8}\right)-7} & & =\frac{24 x-21+21}{24} \\&=\sqrt[3]{x^{3}+\frac{0}{8}} & & =\frac{24 x}{24} \\&=\sqrt[3]{x^{3}} & & =x \\&=x\end{aligned}

Find (fg)(x)( f g ) ( x )(fg)(x) f(x)=−5xg(x)=−8x+6f ( x ) = \sqrt { - 5 x } \quad g ( x ) = \sqrt { - 8 x + 6 }f(x)=−5x​g(x)=−8x+6​

Multiple Choice