Find the inverse of each function - g(x)=- 3/x-1 - 2

Question

Having trouble with problems that have a variable in the denominator. 

Answer 1

Hi there, Malik!

 

The way I always try to teach inverse functions is "swap the x and y variables, then solve for y." Let's try it with another function first, just to demonstrate it:

if f(x) = 3x - 7, let y=f(x). Then

y = 3x - 7

and swapping x and y we get

x = 3y - 7

Now solve for y.

 

x+7 = 3y

(x+7)/3 = y = f^-1(x)

 

And we can check that this is really an inverse by plugging in some values (1, 2, 3) into f(x) and then checking those  results in f^-1(x) to (hopefully) x back.

 

We can do the same thing here!

 

Let's do a different problem first.

 

h(x) = 7/(x + 3) - 1

 

Then swapping x and y, we get

 

x = 7/(y+3) - 1

 

So, let's solve for y! 

 Add 1 to both sides

x+1 = 7/(y+3)

 Multiply both sides by (y+3)

(x+1)(y+3) = 7

 divide both sides by (x+1), which is OK as long as x isn't -1

(y+3)=7/(x+1)

 subtract 3 from both sides, and we get:

y = 7/(x+1) - 3  which should be the inverse.

 

I leave it to you to check it yourself, and I hope you can manage the problem above on your own now!

Answer 2

Assuming the equation is g(x) = -3/(x - 1) - 2

 

Let y = g(x).  Then -3/(x - 1) = y + 2

 

Take the reciprocal of both sides: -(x - 1)/3 = 1/(y + 2)

 

Multiply both sides by -3: x -1 = -3/(y + 2)

 

Finally x  = -3/(y + 2) + 1

 

Swap the values of x and y, then the inverse is then

 

y = g^-1(x) = -3/(x + 2) + 1