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monkey*

  • one year ago

Given f(x)=x^2+7 and g(x)=x-4/x. Find (g o f)(-1). Hint: (g o f)(-1)=g(f(-1))

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  1. freckles
    • one year ago
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    so have you tried finding f(-1) and then finding g(put result of f(-1) here)

  2. freckles
    • one year ago
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    f(-1) tells us to use f(x)=x^2+7 we know to use f(x)=x^2+7 because of the f in f(-1)

  3. freckles
    • one year ago
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    f(x)=x^2+7 we can see here the input is the thing in f( ) example: f(a)=a^2+7 f(a+1)=(a+1)^2+7 f(input)=(input)^2+7 another example; f(2)=2^2+7 can you find f(-1)=?

  4. monkey*
    • one year ago
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    I'm a little lost, sorry.

  5. freckles
    • one year ago
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    ok do you see that f(x)=x^2+7 and you are also to find f(-1)?

  6. freckles
    • one year ago
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    f(x)=x^2+7 since we want to find f(-1) we need to replace those x's with -1

  7. monkey*
    • one year ago
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    Yes.

  8. freckles
    • one year ago
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    so can you find f(-1)?

  9. freckles
    • one year ago
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    |dw:1440459828779:dw|

  10. freckles
    • one year ago
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    let me know when you have found f(-1)

  11. monkey*
    • one year ago
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    Would it be f(-1)-1^2+7?

  12. freckles
    • one year ago
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    well there should be an equal sign somewhere there

  13. freckles
    • one year ago
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    |dw:1440460061483:dw|

  14. freckles
    • one year ago
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    can you simplify (-1)^2+7?

  15. monkey*
    • one year ago
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    8?

  16. freckles
    • one year ago
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    yes f(-1)=(-1)^2+7=(-1)(-1)+7=1+7=8 so f(-1)=8 so f(-1) is 8 now we had to find g(f(-1)) we just found that inside thing the f(-1) thing to be 8 replace f(-1) with 8 you are now really wanting to find g(8)

  17. freckles
    • one year ago
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    g(8) tells us to use the expression named g

  18. freckles
    • one year ago
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    g looks likes g(x)=x-4/x

  19. freckles
    • one year ago
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    just replace the old input x with the new input 8

  20. freckles
    • one year ago
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    simplify and then you are done

  21. monkey*
    • one year ago
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    So, g(8)=x-4/x?

  22. freckles
    • one year ago
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    you have to replace all the x's with 8 not just one of the x's

  23. freckles
    • one year ago
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    whatever you replace the first x with you have to replace the other x's on the other side also

  24. monkey*
    • one year ago
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    g(8)=8-4/8?

  25. freckles
    • one year ago
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    yes

  26. monkey*
    • one year ago
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    So simplify 8-4/8?

  27. monkey*
    • one year ago
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    @freckles

  28. freckles
    • one year ago
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    yes

  29. freckles
    • one year ago
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    I don't think teachers like unsimplified answers :p

  30. monkey*
    • one year ago
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    7.5 is my answer?

  31. monkey*
    • one year ago
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    And no, they don't lol.

  32. monkey*
    • one year ago
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    Thank you so much for the help!

  33. freckles
    • one year ago
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    for me to check your answer I have to know one this is it \[g(x)=\frac{x-4}{x} \text{ or } g(x)=x-\frac{4}{x}\]

  34. monkey*
    • one year ago
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    First one.

  35. freckles
    • one year ago
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    so it is actually g(x)=(x-4)/x not g(x)=x-4/x so g(8)=(8-4)/8=4/8=1/2 or .5

  36. freckles
    • one year ago
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    \[f(x)=x^2+7 \\ g(x)=\frac{x-4}{x} \\ g(f(-1)) = g((-1)^2+7)=g(1+7)=g(8)=\frac{8-4}{8}=\frac{4}{8}=\frac{1}{2}\]

  37. monkey*
    • one year ago
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    So what would be the answer?

  38. freckles
    • one year ago
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    well we just got that g(f(-1))=1/2 and we were looking for the value of g(f(-1)) so we are done

  39. monkey*
    • one year ago
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    So .5 would be the answer?

  40. freckles
    • one year ago
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    yes a half is the answer

  41. freckles
    • one year ago
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    1/2 or .5 you choose the form or write in the desired form your teacher wants

  42. freckles
    • one year ago
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    I like pretty fractions so I would write 1/2 :p

  43. monkey*
    • one year ago
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    Okay, thanks. And she wants a fraction. Could you help with one more?

  44. freckles
    • one year ago
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    sure i can try

  45. monkey*
    • one year ago
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    Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

  46. monkey*
    • one year ago
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    Determine the standard form of the equation of the line that passes through (-7,8) and (0,-8).

  47. freckles
    • one year ago
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    point slope form for a line looks like: \[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line }\]

  48. freckles
    • one year ago
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    you might want to go ahead and find m first

  49. freckles
    • one year ago
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    if you have two points you can find m by using the following formula: \[m=\frac{y_2-y_1}{x_2-x_1}\]

  50. freckles
    • one year ago
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    we do have two points so we can use that formula

  51. monkey*
    • one year ago
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    M=-8-8/0-(-7)?

  52. freckles
    • one year ago
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    \[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\] yes if you mean m=(-8-8)/(0-(-7))

  53. freckles
    • one year ago
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    and yes you simplify this value for m that we just found

  54. freckles
    • one year ago
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    so -8-8=-(8+8)=-16 right and what about the bottom 0-(-7)=-(-7)=...

  55. monkey*
    • one year ago
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    M=-16/7 ?

  56. freckles
    • one year ago
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    bingo! m is -16/7

  57. freckles
    • one year ago
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    \[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\ \] \[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\] so let's go back to our point slope form for a line... \[y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}\] \[y-y_1=\frac{-16}{7}(x-x_1)\]

  58. freckles
    • one year ago
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    now really the last thing to do is put in x1 and y1

  59. freckles
    • one year ago
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    you can use the value I called x1 and y1 above

  60. monkey*
    • one year ago
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    Okay, soo?

  61. freckles
    • one year ago
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    that is all that is really left is to replace x1 and y1 have you done that?

  62. monkey*
    • one year ago
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    No, I'm confused about that.

  63. freckles
    • one year ago
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    " \[y-y_1=m(x-x_1) \\ \text{ where } m \text{ is the slope of the line } \\ \text{ and } (x_1,y_1) \text{ is a point on the line } \\\] \[(x_1,y_1) \text{ and } (x_2,y_2) \\ (-7,8) \text{ and } (0,-8) \\ \text{ so replace } x_1 \text{ with } -7 \\ \text{ and replace } y_1 \text{ with } 8 \\ \text{ replace } x_2 \text{ with } 0 \\ \text{ finally replace } y_2 \text{ with } -8 \\ m=\frac{y_2-y_1}{x_2-x_1}=\frac{-8-8}{0-(-7)}\] so let's go back to our point slope form for a line... \[y-y_1=m(x-x_1) \\ \text{ we just found } m \text{ to be } \frac{-16}{7} \\ \\ \text{ replace } m \text{ with } \frac{-16}{7}\] \[y-y_1=\frac{-16}{7}(x-x_1)\] " You should see somehwere here that I called a number x1 and I called a number y1...

  64. freckles
    • one year ago
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    remember the point (-7,8) ?

  65. freckles
    • one year ago
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    x1 was -7 y1 was 8?

  66. monkey*
    • one year ago
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    Yes! I feel stupid now.

  67. freckles
    • one year ago
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    remember the last thing really is just to replace x1 with -7 and y1 with 8..

  68. freckles
    • one year ago
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    thought this form we are looking at is point-slope form

  69. freckles
    • one year ago
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    and your question asks for standard form

  70. freckles
    • one year ago
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    ax+by=c <--so this is the form we want

  71. monkey*
    • one year ago
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    So, -7x+8y=-16/7 ?

  72. freckles
    • one year ago
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    your line should be \[y-8=\frac{-16}{7}(x-(-7)) \\ \text{ or you could have actually used the point } (x_2,y_2) \text{ instead } \\ y-(-8)=\frac{-16}{7}(x-0) \\ \text{ doesn't matter you just have to plug in a point that is on the line for } \\ (x_1,y_1)\] so either one we play with will give us the same answer... you pick the first one or second one... I honestly think the second one looks prettier

  73. freckles
    • one year ago
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    \[y+8=\frac{-16}{7}x \\ \text{ note: all I did here was do } -(-8)=8 \text{ and } x-0=x \\ \text{ anyways we have so far } y+8=\frac{-16}{7}x \\ \text{ we want our } x \text{ terms and } y \text{ terms on the same side } \\ \text{ so subtract } y \text{ on both sides } 8=\frac{-16}{7}x-y \\ \] in some people's eyes this may be done by some prefer the number in front of x to be positive so you you can multiply both sides by -1 \[-8=\frac{16}{7}x+y\] now some prefer just to also have integers coefficients and 16/7 is definitely not an integer so you could also multiply both sides by 7 \[-56=16x+7y\]

  74. freckles
    • one year ago
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    this is probably the most secure way to write it because it fits everyones' expectation of standard form for a line

  75. freckles
    • one year ago
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    Ax+By=C A is positive integer B is integer C is integer I think this is the most strict definition of a line in standard form

  76. monkey*
    • one year ago
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    I have another one just like this except the points are (9,-4) and (6,4). I do everything the same way, right?

  77. freckles
    • one year ago
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    yep

  78. freckles
    • one year ago
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    put in point slope form then put in standard form

  79. monkey*
    • one year ago
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    Okay, thank you.

  80. freckles
    • one year ago
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    np

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