Question

25 POINTS PLZ HELP ASAP!!! Match the functions with their inverse functions.

Answer 1

Answer with explanation:

We know that when we are given a inverse function as: $f^{-1}(x)$

Then we find the function by the method:

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

Then we switch the places of x and y and solve for y.

1)

$f^{-1}(x)=\dfrac{2x-3}{5}$

Hence, we find the function as follows:

$\dfrac{2x-3}{5}=y$

Then we switch for x and y

$\dfrac{2y-3}{5}=x\\\\\\i.e.\\\\2y-3=5x\\\\\\i.e.\\\\\\2y=5x+3\\\\\\i.e.\\\\\\y=\dfrac{5x+3}{2}$

Hence, inverse function is:

  $f(x)=\dfrac{5x+3}{2}$

2)

$f^{-1}(x)=\dfrac{x+8}{2}$

Hence, we find the function as follows:

$\dfrac{x+8}{2}=y$

Then we switch for x and y

$\dfrac{y+8}{2}=x$

$y=2x-8$

Hence, inverse function is:

           $f(x)=2x-8$

3)

$f^{-1}(x)=\dfrac{x+2}{7}$

Hence, we find the function as follows:

$\dfrac{x+2}{7}=y$

Then we switch for x and y

$\dfrac{y+2}{7}=x$

$y=7x-2$

Hence, inverse function is:

           $f(x)=7x-2$

4)

$f^{-1}(x)=\dfrac{1-2x}{x}$

Hence, we find the function as follows:

$\dfrac{1-2x}{x}=y$

Then we switch for x and y

$\dfrac{1-2y}{y}=x$

i.e.

$1-2y=xy\\\\xy+2y=1\\\\i.e.\\\\y(x+2)=1\\\\y=\dfrac{1}{x+2}$

Hence, inverse function is:

  $f(x)=\dfrac{1}{x+2}$

Answer 2

To calculate the inverse isolate x on one side and switch x with y on both sides

1)
y=(2x-3)/5
5y=2x-3
5y+3=2x
(5y+3)/2=x
-> f(x)=(5x+3)/2

2)
y=(x+8)/2
2y=x+8
2y-8=x
-> f(x)=2x-8

3)
y=(x+2)/7
7y=x+2
7y-2=x
-> f(x)=7x-2

4)
y=(1-2x)/x
y=(1/x)-2
y+2=1/x
x=1/(y+2)
-> f(x)=1/(x+2)