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Here's my problem for today:

If f(x) = {(2,3),(4,5),(7,2)} and g(x) = {(2,1),(3,2),(7,4)}, then determine (f+g)(2) and (f(g(7)).

 

I have no clue how to do this. Help please!

 Jun 26, 2015

Best Answer 

 #1
avatar+129850 
+10

(f + g)(2)  =  f(2)  + g(2)   .........notce that   f(2) means that we are just asking what "y" value is associated with x in the function "f" when x = 2  ......this is the point (2,3)......so  f(2)  = 3

 

Likewise, g(2)   = 1

 

So......  (f + g)(2) = f(2) + g(2)  =  3 +  1 =   4

 

 

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f(g(7)).....this means that, first, we are putting "7" into "g"  an finding the y value associated with that x value.....this is the point (7,4)....so   g(7) = 4

 

So....so far  ......f(g(7))  =  f(4).....now.......we are putting "4" into "f" and finding the y value associated with that x value.......this is the point (4,5)......so f(4)  = 5

 

To recap......f(g(7)  = f(4)  = 5

 

These are known as "composite" functions.....I'll admit they ARE a little tough to grasp !!!!

 

Post some more questions if you get stuck.....!!!

 

 

  

 Jun 26, 2015
 #1
avatar+129850 
+10
Best Answer

(f + g)(2)  =  f(2)  + g(2)   .........notce that   f(2) means that we are just asking what "y" value is associated with x in the function "f" when x = 2  ......this is the point (2,3)......so  f(2)  = 3

 

Likewise, g(2)   = 1

 

So......  (f + g)(2) = f(2) + g(2)  =  3 +  1 =   4

 

 

--------------------------------------------------------------------------------------------------------------------------

f(g(7)).....this means that, first, we are putting "7" into "g"  an finding the y value associated with that x value.....this is the point (7,4)....so   g(7) = 4

 

So....so far  ......f(g(7))  =  f(4).....now.......we are putting "4" into "f" and finding the y value associated with that x value.......this is the point (4,5)......so f(4)  = 5

 

To recap......f(g(7)  = f(4)  = 5

 

These are known as "composite" functions.....I'll admit they ARE a little tough to grasp !!!!

 

Post some more questions if you get stuck.....!!!

 

 

  

CPhill Jun 26, 2015

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