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anonymous

  • one year ago

plz help medal + fan + testimonial show all the work ( explain) do not use a calculator for parts 1-4. Let f be the function defined as follows:

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  1. anonymous
    • one year ago
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    |dw:1435882692544:dw|

  2. anonymous
    • one year ago
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    1. if a=2 and b=3, is f continuous at x=1? justify your answer. 2. find a relationship between a and b for which f is continuous at x=1. ( Hint: A relationship between a and b just means an equation in a and b. 3. Find a relationship between a and b so that f is continuous at x=2. 4. USe your equations from parts (ii)and(III) to find the values of a and b so that f is continuous at both x=1 and at x=2? 5. Graph the piece function using the values of a and b that you have found. you may graph by hand or use your calculator to graph.

  3. Loser66
    • one year ago
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    Any idea?

  4. anonymous
    • one year ago
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    no im completely clueless in this! o.o

  5. Loser66
    • one year ago
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    1) No, since we don't have x =1 on f(x). (The way they define f(x) excludes 1) 2) if f(x) is continuous at x =1, that is the first part = second part (in f(x) I meant if x =1, then y =2 (for f(x) =3-x) Plug it into the second one ax^2 + bx =y with x =1, y =2 to have the relationship between a and b 3) do the same with the second part and the third part for x =2 That is if x =2, y = 0 (for y = 5x-10) Now we have (2,0) , replace in ax^2+bx =y to get the relationship between them.

  6. Loser66
    • one year ago
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    4) combine part 2 and part 3 to solve for a, b. We have 2 equations, 2 unknowns --> the system is solvable. 5) graph, this part is the easiest part. I think you can do it, right?

  7. anonymous
    • one year ago
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    huuuuuuhhhhh? o.ohow did you get that?

  8. Loser66
    • one year ago
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    You asked "how did I get that?" , right? of course, I got it from the courses I took in the past.

  9. anonymous
    • one year ago
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    lol I mean how did you arrive to the answers?

  10. Loser66
    • one year ago
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    I don't get what you mean. My logic is all in what I said. Read and you can see how, why and answer.

  11. anonymous
    • one year ago
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    ok if I need any more help can I message you

  12. anonymous
    • one year ago
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    you got this?

  13. anonymous
    • one year ago
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    is there anything you wold like to add?

  14. anonymous
    • one year ago
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    i was just wondering if you got the answer the question is just walking you through it step by step

  15. anonymous
    • one year ago
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    not really im tryig on my own lol. but I am terrible. even with the steps which I appreciate , I am having a little trouble getting to the answer.

  16. anonymous
    • one year ago
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    this is what you do plug in 1 for \(x\) in both \(3-x\) and \(ax^2+bx\) you get \[3-1=2\] and \[a+b\] setting them equal tells you \[a+b=2\]

  17. anonymous
    • one year ago
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    then plug in \(2\) in \(ax^2+bx\) and \(5x-10\) you get \[4a+2b\] and \[2\times 2-1=0\] telling you \[4a+b=0\]

  18. anonymous
    • one year ago
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    is this for number one?

  19. anonymous
    • one year ago
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    now you have \[a+b=2\\ 4a+2b=0\] solve that for \(a\) and \(b\) and you are done

  20. anonymous
    • one year ago
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    wait do I plug in a+b=2 into 4a+2b=0?

  21. anonymous
    • one year ago
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    you solve the "system of equations" just as normal \[a+b=2\\ 4a+2b=0\] you can solve it using any method you like (or know)

  22. anonymous
    • one year ago
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    ok this is what I get: 4a^2+ 2b^2=2

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