How do you find #(f@g)(x)# given #g(x) = (2x) (1/2)#, #f(x) = x^2 + 1 #?

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Answer 1 · 2016-10-21T16:40:10

See below.

#(f@g)(x)=f(g(x))=g(x)^2-1=x^2-1=f(x)#

Not weird,

because #g(x)=x# is the identity transformation.

Answer 2 · 2016-10-21T17:04:39

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

Let's start by simplifying function #g#.

#g(x) = 2x(1/2) = (2x)/2 = x#

Now, we can rewrite #(f @ g)(x)# as #f(g(x))#.

#f(g(x)) = (x)^2 + 1#

So, #f(g(x)) = f(x)#.

Hopefully this helps!