Show algebraically that the functions

f

and

g

shown below are inverse functions.

f ( x ) = - \frac { 5 } { 7 } x - 3 , \quad g ( x ) = - \frac { 7 x + 21 } { 5 }

A)

\begin{array}{rlrl}f(g(x)) & =-\frac{5}{7}\left(\frac{7 x+21}{5}\right)-3 & g(f(x)) & =-\frac{7\left(-\frac{5}{7} x-3\right)+21}{5} \\& =\left(\frac{7 x+21}{7}\right)-3 & & =-\frac{(-5 x-21)+21}{5} \\& =(x+3)-3 & & =\frac{-5 x-21+21}{5} \\& =x+3-3 & & \\& =x & &= \frac{5 x}{5}\end{array}

B)

\begin{array}{rlrl}f(g(x)) & =-\frac{5}{7}\left(-\frac{7 x+21}{5}\right)-21 & g(f(x)) & =-\frac{7\left(-\frac{5}{7} x-3\right)+21}{5} \\& =\left(\frac{35 x+21}{35}\right)-21 & & =-\frac{(-5 x-3)+21}{5} \\& =(x+21)-21 & & =\frac{5 x+3-21}{5} \\& =x+21-21 & & =\frac{5 x}{5} \\& =x & &=x\end{array}

C)

\begin{array}{l}\begin{array}{c}f(g(x)) &=-\frac{5}{7}\left(-\frac{7 x+3}{5}\right)-3 \\&=\left(\frac{35 x+3}{35}\right)-3 \\&=(x+3)-3 \\&=x+3-3 \\&=x\end{array}\begin{array}{c}g(f(x))=-\frac{7\left(-\frac{5}{7} x-3\right)+21}{5}\\=-\frac{(-5 x-3)+3}{5}\\=\frac{5 x+3-3}{5}\\=\frac{5 x}{5}\\=x\end{array}\end{array}

D)

\begin{array}{l}\begin{array}{lll}f(g(x)) &=-\frac{5}{7}\left(-\frac{7 x+21}{5}\right)-3 \\&=\left(\frac{7 x+21}{7}\right)-3 \\&=(x+3)-3 \\&=x+3-3 \\&=x\end{array}\begin{array}{lll}g(f(x)) &=-\frac{7\left(-\frac{5}{7} x-3\right)+21}{5} \\&=-\frac{(-5 x-21)+21}{5} \\&=\frac{5 x+21-21}{5} \\&=\quad \frac{5 x}{5} \\&=x\end{array}\end{array}

E)

\begin{array}{l}\begin{array}{lll}f(g(x)) &=-\frac{7}{5}\left(-\frac{5 x+15}{7}\right)-3 \\&=\left(\frac{5 x+15}{5}\right)-3 \\&=(x+3)-3 \\&=x+3-3 \\&=x\end{array}\begin{array}{lll}g(f(x)) &=-\frac{7\left(-\frac{5}{7} x-3\right)+21}{5} \\&=-\frac{(-5 x-3)+21}{35} \\&=\frac{5 x+3-21}{35} \\&=\quad \frac{35 x}{35} \\&=\quad x\end{array}\end{array}

Evaluate the Function at the Specified Value of the Independent f(p)=−3p4p−3f ( p ) = \frac { - 3 p } { 4 p - 3 }f(p)=4p−3−3p​

Multiple Choice