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lavalava

  • 3 years ago

use the function to answer the question: x+14 if x<-7 f(x) 21 if -7 ≤x<-6 x^2 -1 if x ≥ -6 what is the value of f(-6)?

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  1. Eyad
    • 3 years ago
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    Since u want to find F(x) at x=-6 , U will use the Function F(x)= x^2-1 because x is bigger or **Equal** -6 Plug -6 in the equation so it will be F(-6)=(-6)^2-1=.....

  2. lavalava
    • 3 years ago
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    35?

  3. hartnn
    • 3 years ago
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    yup,35

  4. Eyad
    • 3 years ago
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    Bingo!

  5. lavalava
    • 3 years ago
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    what about f(x)=x^2-4; g(x)=x+2 which is equal to (f+g)(x)

  6. Eyad
    • 3 years ago
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    U mean Find F(F(x)+g(x)) ?

  7. lavalava
    • 3 years ago
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    no its f(x)+g(x) im not to sure but i think that might be it

  8. hartnn
    • 3 years ago
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    then just add them x^2-4+x+2 = ?

  9. lavalava
    • 3 years ago
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    x^2+x-2?

  10. hartnn
    • 3 years ago
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    yup

  11. lavalava
    • 3 years ago
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    okay so what is f(x)=4x-1 g(x)=5x^2 which is equal to (fºg)(x)?

  12. hartnn
    • 3 years ago
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    for (fog)(x) , replace 'x' in f(x) by g(x) = 5x^2

  13. lavalava
    • 3 years ago
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    so then : 4(5x^2)-1

  14. hartnn
    • 3 years ago
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    yup, that will be 20x^2-1

  15. lavalava
    • 3 years ago
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    and then what is f(x)=4x-7 g(x)=x+3 what is the function of (gºf)(4)

  16. lavalava
    • 3 years ago
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    is it 12?

  17. hartnn
    • 3 years ago
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    to get (gof)(x) replace 'x' in g(x) by f(x) and then put x=4

  18. hartnn
    • 3 years ago
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    yup, its 12

  19. lavalava
    • 3 years ago
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    what is f(x)=5x-10 what is the value of f^-1 (-3)

  20. hartnn
    • 3 years ago
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    http://openstudy.com/study#/updates/50615c4ce4b02e139411444b

  21. hartnn
    • 3 years ago
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    so got that, its 1.4

  22. lavalava
    • 3 years ago
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    yea

  23. hartnn
    • 3 years ago
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    any more questions to come ?

  24. lavalava
    • 3 years ago
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    yea

  25. lavalava
    • 3 years ago
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    what is the vertex of the graph of f(x)=-2|x-5|+3

  26. lavalava
    • 3 years ago
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    i think its -5, 3

  27. Eyad
    • 3 years ago
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    +5,3

  28. hartnn
    • 3 years ago
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    nopes, (5,3) vertex for f(x) = p|x+a|+b is -a,b

  29. lavalava
    • 3 years ago
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    ohhh okay

  30. lavalava
    • 3 years ago
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    the graph of g(x) results when the graph of f(x)=-|x| is shifted 3 units to the left. i think it's g(x)=-|x+3|

  31. lavalava
    • 3 years ago
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    what is the value of f(-4) when f(x)= -10 if x< -5 x^3 if -5≤x≤2 2x+4 if x>2

  32. hartnn
    • 3 years ago
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    in which range does -4 lie ?

  33. lavalava
    • 3 years ago
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    the middle one and the last one

  34. hartnn
    • 3 years ago
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    not the last one, x>2 means x cannot be -4 so only middle one. put x= -4 in middle function, what do u get ?

  35. lavalava
    • 3 years ago
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    -64

  36. hartnn
    • 3 years ago
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    thats your answer :)

  37. lavalava
    • 3 years ago
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    okay now this one is kinda confusing to me but i think i know the answer....

  38. lavalava
    • 3 years ago
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    i think it is the third one, but im not too sure....

  39. hartnn
    • 3 years ago
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    nopes, its 4th. put these values, x=0,0.5,1,1.5,..., find g(x) u will get it

  40. lavalava
    • 3 years ago
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    ohhhh i get it!!! its soo confusing sometimes!

  41. lavalava
    • 3 years ago
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    f(x)=2x+5 g(x)=3x^2 which is equal to (fºg)(x)? i tried it and i got 3x^2+2x+5 but i think it's wrong....

  42. hartnn
    • 3 years ago
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    put x= 3x^2 in f(x) 2(3x^2)+5 = ?

  43. lavalava
    • 3 years ago
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    6x^2+5?

  44. hartnn
    • 3 years ago
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    yup.

  45. lavalava
    • 3 years ago
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    omg you make it easier to understand hahaha

  46. hartnn
    • 3 years ago
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    more questions ? i need to go...

  47. lavalava
    • 3 years ago
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    yea one last one f(x)=2/x g(x)=|x|-1 what is the domain of (fºg)(x)

  48. hartnn
    • 3 years ago
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    fog (x) = ?

  49. lavalava
    • 3 years ago
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    x|x=real numbers x|x ≠0 x|x≠0, x≠1 x|x≠-1, x≠1

  50. hartnn
    • 3 years ago
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    f(g(x)) = 2/(|x|-1) ok? for which values is denominator = 0 ??

  51. lavalava
    • 3 years ago
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    no the ones i put up last were the options

  52. hartnn
    • 3 years ago
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    i know. denominator should not be 0 |x|-1 should not be 0 for x=1, |1|-1=1-1=0 for x=-1, |-1|-1=1-1=0 so x|x≠-1, x≠1 got it ?

  53. lavalava
    • 3 years ago
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    ohh okay yea! thank you! you make it easier to understand ty very much

  54. hartnn
    • 3 years ago
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    welcome :)

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