Answer:
[tex]f^{-1}[/tex](x) = [tex]\frac{2-x}{x-1}[/tex]
Step-by-step explanation:
Let y = f(x) and rearrange making x the subject
y = [tex]\frac{x+2}{x+1}[/tex] ( multiply both sides by (x + 1)
y(x + 1) = x + 2 ← distribute left side
yx + y = x + 2 ( subtract x from both sides )
yx - x + y = 2 ( subtract y from both sides )
yx - x = 2 - y ← factor out x on the left side
x(y - 1) = 2 - y ← divide both sides by (y - 1 )
x = [tex]\frac{2-y}{y-1}[/tex]
Change y back into terms of x, thus
[tex]f^{-1}[/tex](x) = [tex]\frac{2-x}{x-1}[/tex]
Answer: [tex]\bold{f^{-1}(x)=\dfrac{2-x}{x-1}\qquad \implies \qquad f^{-1}(x)=-\dfrac{x-2}{x-1}}[/tex]
Step-by-step explanation:
Inverse is when you swap the x's and y's and solve for y. f(x) is y.
[tex]y=\dfrac{x+2}{x+1}\\\\\\\\\underline{\text{Swap the x's and y's}}\\\\x=\dfrac{y+2}{y+1}\\\\\\\\\underline{\text{solve for y}}\\\\x(y+1)=y+2\\\\xy+x=y+2\qquad \qquad \text{distributed x into y+1}\\\\xy-y=2-x\qquad \qquad \text{subtracted x and y from both sides}\\\\y(x-1)=2-x\qquad \qquad \text{factored out y on the left side}\\\\\\y=\dfrac{2-x}{x-1}}\qquad \qquad \text{divided both sides by x-1}[/tex]
[tex]\large\boxed{f^{-1}(x)=\dfrac{2-x}{x-1}}[/tex]
This is equal to: [tex]f^{-1}(x)=-\dfrac{x-2}{x-1}[/tex]
