# For the function #g(x) = 32/(x+3)#, what is #(g^-1 cdot g)(5)#?

##### 2 Answers

#### Answer:

#### Explanation:

Let's start by finding an equation for the inverse.

#y = 32/(x+ 3)#

Switch the x and the y.

#x = 32/(y + 3)#

#x(y + 3) = 32#

#xy + 3x = 32#

#xy = 32 - 3x#

#y = (32 - 3x)/x#

Now let's complete the required operation.

#(g^-1 * g)(5) = (32 - 3x)/x * 32/(x + 3)#

#(g^-1 * g)(5) = (32 - 15)/5 * 32/(5 + 3)#

#(g^-1 * g)(5) = 17/5 * 4#

#(g^-1 * g)(5) = 68/5#

Hopefully this helps!

#### Answer:

# (g^(-1)*g)(5) = 5 #

#### Explanation:

It should be obvious that irrespective of the definition of the function

# (g^(-1)*g)(x) -= x #

That is, in words, if we apply any function to a variable

If you are not convinced, then we can easily demonstrate this by determining an expression for the inverse function

# g(x)=32/(x+3) => g=32/(x+3) #

# :. x+3=32/g#

# :. x=32/g-3#

So for the function

# g^(-1)(x)=32/x-3#

Now let us find

# g(5)=32/(5+3) = 32/8#

Now let us find the composition

# (g^(-1)*g)(5) = g^(-1)(g(5)) #

# " " = g^(-1)(32/8) #

# " " = 32/(32/8)-3 #

# " " = 8-3 #

# " " = 5 #