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)=(x+20)/(x-5)`

Let's solve for the following:

• (a). The inverse of the function.

Rewrite f(x) for y

`y=(x+20)/(x-5)`

Interchange the variables:

`x=(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} (y(x-1))/(x-1)=(20+5x)/(x-1) \\ \\ y=(20+5x)/(x-1) \end{gathered}`

Therefore, the inverse of the function:

`f^(-1)(x)=(20+5x)/(x-1)\text{ for x }\\e\text{ 1}`

Part B.

Now, let's verify the following:

`\begin{gathered} f(f^(-1)(x))=x \\ \\ f((20+5x)/(x-1))=((20+5x)/(x-1)+20)/((20+5x)/(x-1)-5) \\ \\ =(5((4+x)/(x-1))+5(4))/(5((4+x)/(x-1))-1) \\ \\ Cancel\text{ all like terms} \\ \\ =((4+x)/(x-1)+4)/((4+x)/(x-1)-1) \\ \\ =(4+x+4(x-1))/(4+x-1(x-1)) \\ \\ =(4+x+4x-4)/(4+x-x+1) \\ \\ =(5x-4+4)/(4+1) \\ \\ =(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)((x+20)/(x-5)) \\ \\ \text{ Substitute }(x+20)/(x-5)\text{ for x in }f^(-1)(x): \\ \\ f^(-1)(f(x))=(20+5((x+20)/(x-5)))/(((x+20)/(x-5))-1) \\ \\ Multiply\text{ all expressions by \lparen x-5\rparen:} \\ =(20(x-5)+5(x+20))/((x+20)-1(x-5)) \\ \\ =(20x-20(5)+5x+20(5))/(x+20-x+5) \\ \\ =(20x-100+5x+100)/(x-x+20+5) \\ \\ =(20x+5x-100+100)/(x-x+20+5) \\ \\ =(25x)/(25) \\ \\ =x \end{gathered}`

Therefore, we have:

`f^(-1)(f(x))=x`

ANSWER:

Part A: C

`f^(-1)(x)=(20+5x)/(x-1)\text{ for x}\\e1`