Let f(x)=2/x-3. Find f^-1(x)

The inverse when graphed is a reflection through the y=x line. The ordered pairs [tex](a,b)[/tex] on [tex]f[/tex] implies [tex](b,a)[/tex] are on [tex]f^{-1}[/tex].

This means we really just need to swap x and y.

Since we want to write as a function of x we will need to solve for y again.

Question 1:

[tex]y=\frac{2}{x}-3[/tex]

Swap x and y:

[tex]x=\frac{2}{y}-3[/tex]

We want to solve for y.

Add 3 on both sides:

[tex]x+3=\frac{2}{y}[/tex]

Make the left hand side a fraction so we can cross-multiply:

[tex]\frac{x+3}{1}=\frac{2}{y}[/tex]

Cross multiply:

[tex]y(x+3)=1(2)[/tex]

Simplify right hand side:

[tex]y(x+3)=2[/tex]

Divide both sides by (x+3):

[tex]y=\frac{2}{x+3}[/tex]

So [tex]f^{-1}(x)=\frac{2}{x+3}[/tex].

Question 2:

[tex]y=\frac{2}{x-3}[/tex]

Swap x and y:

[tex]x=\frac{2}{y-3}[/tex]

Make left hand side a fraction so we can cross multiply:

[tex]\frac{x}{1}=\frac{2}{y-3}[/tex]

Cross multiply:

[tex](y-3)x=1(2)[/tex]

We have to distribute here:

[tex]yx-3x=2[/tex]

Add 3x on both sides:

[tex]yx=2+3x[/tex]

Divide boht sides by x:

[tex]y=\frac{2+3x}{x}[/tex]

You could probably stop here but you could also simplify a little.

Separate the fraction into two terms since you have 2 terms on top bottom being dividing by x:

[tex]y=\frac{2}{x}+\frac{3x}{x}[/tex]

Simplify second fraction x/x=1:

[tex]y=\frac{2}{x}+3[/tex]

So [tex]f^{-1}(x)=\frac{2}{x}+3[/tex].