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anonymous

  • one year ago

Just checking. If ƒ(x ) = x 2 + 1 and g(x ) = 3x + 1, find 2ƒ(1) + 3g(4).

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  1. anonymous
    • one year ago
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    I came up with 42. Is that correct?

  2. beginnersmind
    • one year ago
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    That's not what I got. What did you get for 2f(1) and 3g(4)?

  3. anonymous
    • one year ago
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    5 and 37

  4. beginnersmind
    • one year ago
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    Hm. 2f(1) is 2*(f(1)) or 2 times the value of the function f(x) at 1.

  5. beginnersmind
    • one year ago
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    So you should start by calculating f(1) and g(4), by plugging x=1 into f(x) and x=4 into g(x). Is that what you did?

  6. anonymous
    • one year ago
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    Yes.

  7. anonymous
    • one year ago
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    What was your final answer? Maybe I can plug it in and see if it is a match.

  8. beginnersmind
    • one year ago
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    The final answer is 43. f(1) = 2 2f(1) =4 g(4) = 13 3g(4) = 39

  9. beginnersmind
    • one year ago
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    I'm guessing x2 meant x^2 in your problem, not 2x.

  10. anonymous
    • one year ago
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    right. did you do 13*3

  11. beginnersmind
    • one year ago
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    Yes.

  12. anonymous
    • one year ago
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    I think I got it thanks.

  13. beginnersmind
    • one year ago
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    No problem. If you need to go through a few more I'll be happy to help.

  14. anonymous
    • one year ago
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    Alright. If ƒ(x ) = x^2 + 1 and g(x ) = 3x + 1, find [ƒ(2) - g(1)]^2.

  15. beginnersmind
    • one year ago
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    Alright, start by calculating f(2) and g(1)

  16. anonymous
    • one year ago
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    5 and 4

  17. beginnersmind
    • one year ago
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    |dw:1441895933081:dw|

  18. beginnersmind
    • one year ago
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    right

  19. anonymous
    • one year ago
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    when does the exponent outside of the bracket come into play

  20. anonymous
    • one year ago
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    5-4=1 1^2 is 1

  21. beginnersmind
    • one year ago
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    Correct. The exponent "comes into play" when you calculate what's the value of the base. (that is the expression inside the [ ] parentheses.)

  22. anonymous
    • one year ago
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    How about If ƒ(x ) = x^2 + 1, find ƒ(a + 1).

  23. anonymous
    • one year ago
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    (a+1)^2 + 1

  24. beginnersmind
    • one year ago
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    That's essentially correct.

  25. beginnersmind
    • one year ago
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    You can write it out as a^2 + 2a + 2, if you want.

  26. anonymous
    • one year ago
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    Given the function ƒ(x ) = 3x + 1, evaluate ƒ(a + 1) Could you use distributive So 3(a+1)+1 becomes 3a+3+1 then 3a+4

  27. beginnersmind
    • one year ago
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    Yes.

  28. anonymous
    • one year ago
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    Great last one.If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) =

  29. anonymous
    • one year ago
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    3h

  30. beginnersmind
    • one year ago
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    After you get the result f(a+1) = 3(a+1) + 1 you can treat the result as any other expression.

  31. beginnersmind
    • one year ago
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    Great last one.If ƒ(x ) = 3x + 1, then ƒ(a + h ) - ƒ(a ) = 3h correct

  32. anonymous
    • one year ago
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    Thank you. You have been such a great help.

  33. beginnersmind
    • one year ago
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    Cool :)

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