How do you find #(f @ g)(x)# and its domain, #(g @ f)(x)# and its domain,# (f @ g)(2) # and #(g @ f)(2)# of the following problem #f(x) = x+ 2#, #g(x) = 2x^2#?
To find ( (f @ g)(x) ), ( (g @ f)(x) ), ( (f @ g)(2) ), and ( (g @ f)(2) ) for the functions ( f(x) = x + 2 ) and ( g(x) = 2x^2 ), follow these steps:

( (f @ g)(x) ) is obtained by substituting ( g(x) ) into ( f(x) ). So, ( (f @ g)(x) = f(g(x)) = f(2x^2) = 2x^2 + 2 ).
The domain of ( (f @ g)(x) ) is the set of all real numbers since there are no restrictions on the domain of ( 2x^2 + 2 ).

( (g @ f)(x) ) is obtained by substituting ( f(x) ) into ( g(x) ). So, ( (g @ f)(x) = g(f(x)) = g(x + 2) = 2(x + 2)^2 = 2(x^2 + 4x + 4) = 2x^2 + 8x + 8 ).
The domain of ( (g @ f)(x) ) is also the set of all real numbers.

To find ( (f @ g)(2) ), substitute ( 2 ) into ( 2x^2 + 2 ): ( (f @ g)(2) = 2(2)^2 + 2 = 2(4) + 2 = 10 ).

To find ( (g @ f)(2) ), substitute ( 2 ) into ( 2x^2 + 8x + 8 ): ( (g @ f)(2) = 2(2)^2 + 8(2) + 8 = 2(4)  16 + 8 = 16  16 + 8 = 8 ).
So, ( (f @ g)(2) = 10 ) and ( (g @ f)(2) = 8 ).
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See the explanation below...
A function's domain is the range of values that the function is defined and real for.
Likewise, we have:
Simplify:
Simplify:
And:
Simplify:
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When evaluating a onesided 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.
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