Question

Can someone please help me with this math problem, ASAP?

Answer 1

`\mbox{\large s^(-1)(x)} =\mbox{\large \boxed{(2-2x)/(3x)}}`

Explanation:

Given function is

`s(x) = (2)/(3x + 2)\\\\`

To find the inverse for convenience, let y = s(x)

`\rightarrow\quad y = (2)/(3x+2)\\\\`

Replace x with y:

`\rightarrow\quad x = (2)/(3y+2)`

Solve for y:

• Multiply both sides by 3y + 2
  
  `x\left(3y+2\right)=(2)/(3y+2)\left(3y+2\right)`
  
• Simplify
  
  `x\left(3y+2\right)=2`
  
• Divide both sides by x and simplify
  
  `(x\left(3y+2\right))/(x)=(2)/(x)\\\\3y+2=(2)/(x)`
  
• Move 2 to the right side
  
  `3y = (2)/(x) - 2; \\\\3y = (2-2x)/(x)\\`
  
• Divide both sides by 3:
  
  `y=(2-2x)/(3x)`

• Replace y with
  `s^(-1)(x)`
  
  `s^(-1)(x) = y=(2-2x)/(3x)`