Question #715a8

1 Answer
Peter M.
Jun 5, 2017

#(f@g)(x)=3x^2-27#

Explanation:

This is a composition of functions. It is essentially a function within a function. The notation states that you are evaluating the inner function, g(x), and then inserting the result into f(x) and re-evaluating.

#(f@g)(x)=f(g(x))#

Read it as "f of g". I prefer the second notation as you can see which one is inside the other. #f(x)# is the outer function and #g(x)# is the inner function.

To evaluate, you replace all the #x#'s of the outer function with the inner function.

#f(x)=x -4#
#g(x)=3x^2-23#

#(f@g)(x)=f(g(x))=f(3x^2-23)#

#=(3x^2-23)-4#

#=3x^2-27#

As an example, let's evaluate #(f@g)(x)# when #x=2#. You can do this by step by step:

i) Evaluate the inner function

#g(2)=3(2)^2-23=-11#

ii) Insert this into the outer function and evaluate

#f(-11)=-11 -4=-15#

But now that we have the composition function, we can insert #x=2# directly into it instead to get the same answer:

#(f@g)(2)=3(2)^2-27=-15#

*As extra but vital information, an important rule for the composition to be defined is this:

#"range of inner function " sube " domain of outer function"#
I.e.
#"ran " g(x) sube "dom "f(x)#

This says that the range of #g(x)# must be a subset or equal to the domain of #f(x)#. If this is not true, you trying to input values into #f(x)# that are not part of its domain and, thus, cannot be evaluated.
In this case,

#"ran " g(x) = [-23,oo)#
#"dom "f(x)=RR#

The below statement is true, so the composition is defined.

#[-23,oo) sube RR#

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