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Assessment 01.03 Essay

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Assessment 01.03 Module One Quiz.

1. Splott and Fizzle were solving for 2x+4=-3x+14. Both of them made some errors like when 2x+4=-3x+14 what they should have done was change the sign from -3x to +3x and moved it to the other side with the 2x, put both variables together, which would have been (2x+3x) +4= (+3x where the –3x was which is canceled out) +14, when solved it would have been 5x+4=14, then you would do the opposite of +4 to -4 to its original (the variable number is on one side with the equal sign) and subtracted from the 14 so It would come out to 5x=14-4 you solve 5x=10, divide by 5, x=10 over 5 solve for x, it would be 2
2. The function I created would be f(x) = 6x+11
3. F(x) =6x=11 is a legitimate function because if you graph y=6x=11 and put a line through it, it would only have one point on (0,11).
4. To solve for f(3) the problem would be f(3)=6(3)+11 to get f(3) you would multiply the 6 with the (3) =18 plus the 11 so f(3)=29
5. To find the inverse f(x) =6x+11 first you put y where the f(x) is then you swap the y for the x that’s with the 6 so it would be x=6y+11 subtract 11 from its self and from the x you end up with x-11=6y you divide the 6y by 6 to get the y alone you also divide the 6 from the 11 that’s when you get your answer x-11/6=y put the f-1(x) where the y is so x-11/6=f-1(x)
6. Using the function I created, If you put any number to x it would come out the same answer because when you do f(g(x)) and g(f(x)) they would not be the same answer at all

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