Question

The function f(x) =x + 20x-5is one-to-one. For the function,a. Find an equation for f (x), the inverse function.b. Verify that your equation is correct by showing that f(f¯¹(x))(f(x)) = x.

Answer 1

Given the function:

$f(x)=\frac{x+20}{x-5}$

Let's solve for the following:

• (a). The inverse of the function.

Rewrite f(x) for y

$y=\frac{x+20}{x-5}$

Interchange the variables:

$x=\frac{y+20}{y-5}$

Solve for y:

$\begin{gathered} x(y-5)=y+20 \\ \\ xy-5x=y+20 \\ \\ xy-5x-y=20 \\ \\ xy-y=20+5x \end{gathered}$

Factor the left side:

$y(x-1)=20+5x$

Divide both sides by (x - 1):

$\begin{gathered} \frac{y(x-1)}{x-1}=\frac{20+5x}{x-1} \\ \\ y=\frac{20+5x}{x-1} \end{gathered}$

Therefore, the inverse of the function:

$f^{-1}(x)=\frac{20+5x}{x-1}\text{ for x }\ne\text{ 1}$

Part B.

Now, let's verify the following:

$\begin{gathered} f(f^{-1}(x))=x \\ \\ f(\frac{20+5x}{x-1})=\frac{\frac{20+5x}{x-1}+20}{\frac{20+5x}{x-1}-5} \\ \\ =\frac{5(\frac{4+x}{x-1})+5(4)}{5(\frac{4+x}{x-1})-1} \\ \\ Cancel\text{ all like terms} \\ \\ =\frac{\frac{4+x}{x-1}+4}{\frac{4+x}{x-1}-1} \\ \\ =\frac{4+x+4(x-1)}{4+x-1(x-1)} \\ \\ =\frac{4+x+4x-4}{4+x-x+1} \\ \\ =\frac{5x-4+4}{4+1} \\ \\ =\frac{5x}{5} \\ \\ =x \end{gathered}$

Therefore, we have:

$f(f^{-1}(x))=x$

Also, let's verify f¯¹(f(x)) = x:

$\begin{gathered} f^{-1}(f(x))=f^{-1}(\frac{x+20}{x-5}) \\ \\ \text{ Substitute }\frac{x+20}{x-5}\text{ for x in }f^{-1}(x): \\ \\ f^{-1}(f(x))=\frac{20+5(\frac{x+20}{x-5})}{(\frac{x+20}{x-5})-1} \\ \\ Multiply\text{ all expressions by \lparen x-5\rparen:} \\ =\frac{20(x-5)+5(x+20)}{(x+20)-1(x-5)} \\ \\ =\frac{20x-20(5)+5x+20(5)}{x+20-x+5} \\ \\ =\frac{20x-100+5x+100}{x-x+20+5} \\ \\ =\frac{20x+5x-100+100}{x-x+20+5} \\ \\ =\frac{25x}{25} \\ \\ =x \end{gathered}$

Therefore, we have:

$f^{-1}(f(x))=x$

ANSWER:

Part A: C

$f^{-1}(x)=\frac{20+5x}{x-1}\text{ for x}\ne1$