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Assessment 01.03 Essay

  • Submitted by: ccdqaf13
  • on February 17, 2015
  • Category: Miscellaneous
  • Length: 353 words

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Below is an essay on "Assessment 01.03" from Anti Essays, your source for research papers, essays, and term paper examples.

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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