Question

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

Answer 1

Answer:

Two questions:

Question 1: $f^{-1}(x)=?$ given $f(x)=\frac{2}{x}-3$.

Answer 1: $f^{-1}(x)=\frac{2}{x+3}$

Question 2: $f^{-1}(x)=?$ given $f(x)=\frac{2}{x-3}$.

Answer 2: $f^{-1}(x)=\frac{2}{x}+3$

Step-by-step explanation:

So $f^{-1}$ is used in most classes to represent the inverse function of $f$.

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

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:

$y=\frac{2}{x}-3$

Swap x and y:

$x=\frac{2}{y}-3$

We want to solve for y.

Add 3 on both sides:

$x+3=\frac{2}{y}$

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

$\frac{x+3}{1}=\frac{2}{y}$

Cross multiply:

$y(x+3)=1(2)$

Simplify right hand side:

$y(x+3)=2$

Divide both sides by (x+3):

$y=\frac{2}{x+3}$

So $f^{-1}(x)=\frac{2}{x+3}$.

Question 2:

$y=\frac{2}{x-3}$

Swap x and y:

$x=\frac{2}{y-3}$

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

$\frac{x}{1}=\frac{2}{y-3}$

Cross multiply:

$(y-3)x=1(2)$

We have to distribute here:

$yx-3x=2$

Add 3x on both sides:

$yx=2+3x$

Divide boht sides by x:

$y=\frac{2+3x}{x}$

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:

$y=\frac{2}{x}+\frac{3x}{x}$

Simplify second fraction x/x=1:

$y=\frac{2}{x}+3$

So $f^{-1}(x)=\frac{2}{x}+3$.