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If #f(x) = x + 3# and #g(x) = 2x - 7#, what is #(f@g)(x)#?

Answer 1

#(f@g)(x)# is equivalent to #f (g(x))#. We solve this problem just as we solve #f(x)#. But since it asks us to find out #f(g(x))#, in #f(x)#, each time we encounter x, we replace it with #g(x)#. In the above problem, #f(x) = x+3#. Therefore, #f(g(x))= g(x) +3#. #=> (f@g)(x)= 2x - 7 + 3# #=> (f@g)(x)= 2x - 4#

Basically, write the #g(x)# equation where you see the x in the #f(x)# equation.

#f@g(x)= (g(x))+3# Replace #g(x)# with the equation #f@g(x)= (2x-7)+3# #f@g(x)= 2x-7+3# we just took away the parentheses #f@g(x)= 2x-4# Because the #-7+3=4# This is it #g@f(x)# would be the other way around

#g@f(x)= 2(x+3)-7#

now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them. #g@f(x)= 2x+6-7#

Next, #+6-7 = -1#

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

Its a lt easier than you think!

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Answer 2

Madelyn Anderson

To find ( (f@g)(x) ), first find ( f(g(x)) ).

( g(x) = 2x - 7 )

( f(g(x)) = f(2x - 7) )

( f(x) = x + 3 )

( f(g(x)) = 2x - 7 + 3 = 2x - 4 )

So, ( (f@g)(x) = 2x - 4 ).

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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.