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

Answer 1

To find ( (f@g)(x) ) given ( g(x) = \sqrt{2x} ) and ( f(x) = x^2 + 1 ), we substitute ( g(x) ) into ( f(x) ), which means replacing ( x ) in ( f(x) ) with ( g(x) ).

[ f@g(x) = f(g(x)) = f(\sqrt{2x}) = (\sqrt{2x})^2 + 1 = 2x + 1 ]

So, ( (f@g)(x) = 2x + 1 ).

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Answer 2

Sophie Adams

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.

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Answer 3

Elijah Allen

#(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!

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.