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

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Answer 1 · 2017-07-01T15:54:56

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

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 2 · 2017-07-01T21:49:48

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

It should be obvious that irrespective of the definition of the function #g# then the composition:

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

That is, in words, if we apply any function to a variable #x#, then apply the inverse function the we end up with what we started with, i.e. #x#

If you are not convinced, then we can easily demonstrate this by determining an expression for the inverse function #g^(-1)(x)#:

# g(x)=32/(x+3) => g=32/(x+3) #
# :. x+3=32/g#
# :. x=32/g-3#

So for the function #g(x)=32/(x+3)#, the inverse function #g^(-1)(x)# is defined by:

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

Now let us find #g(5)#

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

Now let us find the composition #(g^(-1)*g)(5)#

# (g^(-1)*g)(5) = g^(-1)(g(5)) #
# " " = g^(-1)(32/8) #
# " " = 32/(32/8)-3 #
# " " = 8-3 #
# " " = 5 #