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heyitslizzy13

  • one year ago

can someone help me with a problem please!!

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  1. heyitslizzy13
    • one year ago
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  2. heyitslizzy13
    • one year ago
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    @satellite73 @Directrix @jim_thompson5910

  3. Jhannybean
    • one year ago
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    which part of the problem?

  4. heyitslizzy13
    • one year ago
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    The first part for now :)

  5. freckles
    • one year ago
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    Hint: Intermediate Value theorem

  6. heyitslizzy13
    • one year ago
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    @freckles do you know any good websites to review the theorem?

  7. Jhannybean
    • one year ago
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    \[f(x) = ax^3-11x^2+bx+c \qquad f(-3)=10, \qquad f(0)=4,\qquad f(-1)=4\]\[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10 \]\[f(-3)=-27a-99-3b+c=10\]\[f(-3) = -27a-3b+c=109\] ...perhaps we can solve this as a system of equations?

  8. heyitslizzy13
    • one year ago
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    could you show me how? please and thank you :)

  9. jim_thompson5910
    • one year ago
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    part a) imagine you have a river that is a complete straight line |dw:1443850668196:dw| this straight line would stretch forever in both directions

  10. jim_thompson5910
    • one year ago
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    now let's say we had point A and point B one point on either side |dw:1443850712927:dw|

  11. jim_thompson5910
    • one year ago
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    if we connect the two points with a continuous function, then it is impossible to not pass through the line. we might have something like this |dw:1443850761686:dw| the path has to cross over the river somewhere

  12. heyitslizzy13
    • one year ago
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    okay

  13. jim_thompson5910
    • one year ago
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    now imagine the river is the x axis if point A has a positive y coordinate (ie it's above the x axis) and point B has a negative y coordinate (ie it's below the x axis) then the continuous function must cross the x axis at some point between A and B. So if we know A is above the x axis and B is below, there is at least one root in between

  14. Jhannybean
    • one year ago
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    \[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c=4\]\[f(-1) =-a-b+c=15\] \[\therefore \]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\] As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken.

  15. heyitslizzy13
    • one year ago
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    I am kind of confused, how do we know that c=4 is the root that lies in the middle?

  16. Jhannybean
    • one year ago
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    because when \(x=0\) (that means it's at the origin) \(y=4\)

  17. heyitslizzy13
    • one year ago
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    ohhh okay

  18. jim_thompson5910
    • one year ago
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    Jhannybean plugged in x = 0 and everything went away basically except for that last term

  19. Jhannybean
    • one year ago
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    Exactly

  20. heyitslizzy13
    • one year ago
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    do you understand what part a of the problem is asking? I'm quite confused

  21. Jhannybean
    • one year ago
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    Okay so between \(x=-3\) and \(x=0\), is there ANYWHERE on the x-axis when the function gives you a solid number? hint: look at the above posts and our explanations.

  22. jim_thompson5910
    • one year ago
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    if you need a visual way to approach this, draw a blank xy axis |dw:1443851396111:dw|

  23. heyitslizzy13
    • one year ago
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    at -1 i believe , (-1, 4) ?

  24. jim_thompson5910
    • one year ago
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    then plot the point (-3,10), from the fact that f(-3) = 10 and the point (0,4), because f(0) = 4 |dw:1443851450292:dw|

  25. jim_thompson5910
    • one year ago
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    now, ask yourself this: is it possible to connect the two points without crossing the x axis?

  26. heyitslizzy13
    • one year ago
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    it is possible

  27. jim_thompson5910
    • one year ago
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    so there is no guarantee we have a root on the interval from x = -3 to x = 0

  28. jim_thompson5910
    • one year ago
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    now IF we had (0,-4) instead, then we shift down the second point |dw:1443851620081:dw|

  29. jim_thompson5910
    • one year ago
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    |dw:1443851638079:dw|

  30. jim_thompson5910
    • one year ago
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    now it's impossible to connect the two points without crossing the x axis so there has to be at least one root between the two x values |dw:1443851682979:dw| or |dw:1443851692524:dw| or |dw:1443851699567:dw|

  31. heyitslizzy13
    • one year ago
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    ohhh okay!

  32. jim_thompson5910
    • one year ago
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    so really the only guarantee that we have a root is if there is a sign change (in my example the sign change goes from y = 10 to y = -4)

  33. jim_thompson5910
    • one year ago
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    sign change in the y values or function outputs I mean

  34. jim_thompson5910
    • one year ago
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    but since y = 10 to y = 4 has no sign changes in the problem, there is no guarantee there's a root between x = -3 to x = 0

  35. heyitslizzy13
    • one year ago
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    oh i see now

  36. heyitslizzy13
    • one year ago
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    could you help me on part b?

  37. jim_thompson5910
    • one year ago
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    ok so I think you found that c = 4 right?

  38. jim_thompson5910
    • one year ago
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    do you see how Jhannybean got this? \[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\]

  39. heyitslizzy13
    • one year ago
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    yes :)

  40. heyitslizzy13
    • one year ago
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    wait i don't understand how she got the 109, 4 and15

  41. jim_thompson5910
    • one year ago
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    I think f(-3) should have been 10 and not 109

  42. heyitslizzy13
    • one year ago
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    okay, and how is f(-1)= 15?

  43. jim_thompson5910
    • one year ago
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    it should be \[f(-3) = -27a-3b+c=10\\ f(0) =c=4 \\f(-1)=-a-b+c=4\]

  44. heyitslizzy13
    • one year ago
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    oh okay, i understand then

  45. jim_thompson5910
    • one year ago
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    oh wait, there's an 11 in there that I missed, let me rethink

  46. jim_thompson5910
    • one year ago
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    Jhannybean has it right, I missed a term

  47. Jhannybean
    • one year ago
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    \[f(-3)=(-3)^3a -11(-3)^2+(-3)b+c=10\]\[f(-3) = -27a -99 -3b +c = 10\]\[f(-3) = -27a -99 -3b + c +99 = 10 + 99\]\[f(-3) = -27a -3b+c=109\]

  48. Jhannybean
    • one year ago
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    \(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean \[f(0) = a(0)^3 -11(0)+b(0)+c =4\]\[f(0) = 0+0+0+c=4 \\ f(0)=c=4\] \[f(-1) = a(-1)^3 -11(-1)^2+b(-1)+c=4\]\[f(-1) = -a-11-b+c+11=4+11\]\[f(-1) =-a-b+c=15\] \[\therefore\]\[f(-3) = -27a-3b+c=109 \\ f(0) =c=4 \\f(-1)=-a-b+c=15\] As @jim_thompson5910 stated, there is a root that lies in the middle, and from our calculations, that root would be \(c=4\) if i'm not mistaken. \(\color{#0cbb34}{\text{End of Quote}}\)

  49. heyitslizzy13
    • one year ago
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    oh ok

  50. Jhannybean
    • one year ago
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    Does that make more sense, @heyitslizzy13 ?

  51. heyitslizzy13
    • one year ago
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    yes :)

  52. Jhannybean
    • one year ago
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    Great \(\checkmark\)

  53. jim_thompson5910
    • one year ago
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    `that root would be c=4 ` I don't know about that part

  54. heyitslizzy13
    • one year ago
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    so we know c is 4, what about a and b?

  55. Jhannybean
    • one year ago
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    Haha, I meant \(x=0\) x_x

  56. jim_thompson5910
    • one year ago
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    ah I gotcha

  57. heyitslizzy13
    • one year ago
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    would it be 109 and 15?

  58. jim_thompson5910
    • one year ago
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    since c = 4 we can replace each 'c' with 4 \[-27a-3b+\color{red}{c}=109\\-a-b+\color{red}{c}=15\] \[-27a-3b+\color{red}{4}=109\\-a-b+\color{red}{4}=15\]

  59. Jhannybean
    • one year ago
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    So then you would solve it either by substitution, or using the elimination method :)

  60. heyitslizzy13
    • one year ago
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    can you walk me through either one please? I know what they are but I'm a little lost rn

  61. Jhannybean
    • one year ago
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    First we want to isolate a and b in `both` equations to one side, therefore we need to start by subtracting -4 to both sides of both the equations. \[-27a-3b+4-4=109 -4\]\[-a-b+4-4=15-4\] Simplifying both of these, we get: \[-27a-3b=105\]\[-a-b=11\] With me so far?

  62. heyitslizzy13
    • one year ago
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    yes!

  63. Jhannybean
    • one year ago
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    So I personally like using the elimination method. How that works is we're trying to eliminate one of the variables by using a multiplier so when we add the two equations together, we'll be left with one variable, then we can use that to solve for the other. Here, i'm going to eliminate \(b\). in order to eliminate \(b\), i need to multiply the second equation by \(-3\). \[-27a-3b=105\]\[\color{red}{-3}(-a-b=11)\] \[-27a-3b=105\]\[\color{red}{+3}a\color{red}{+3}b=\color{red}{-33}\] Adding these two equations together, we will get \[(-27+3)a +(-3+3)b =105-33\]\[-24a+0b=72\]Now we just solve for \(a\).

  64. Jhannybean
    • one year ago
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    Is this making sense, @heyitslizzy13 ?

  65. heyitslizzy13
    • one year ago
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    yes!! :)) is a=-3?

  66. Jhannybean
    • one year ago
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    Yes it is :)

  67. Jhannybean
    • one year ago
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    Now we just plug \(a=-3\) back into one of the two equations and we can solve for \(b\) I'm going to plug it into \(-a-b=11\) . \[-(-3)-b=11\]\[3-b=11\]\[-3+3-b=11-3\]\[-b=8\]\[b=-8\]

  68. heyitslizzy13
    • one year ago
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    okay thanks :)))

  69. heyitslizzy13
    • one year ago
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    can you help me on c also?

  70. jim_thompson5910
    • one year ago
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    for part c) I would graph the function with a graphing calculator if you don't have one, use something like desmos https://www.desmos.com/calculator

  71. heyitslizzy13
    • one year ago
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    thanks!!

  72. jim_thompson5910
    • one year ago
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    the cool thing about desmos, as compared to a TI83, is that you can use sliders to change the graph https://www.desmos.com/calculator/tvrsgalven

  73. heyitslizzy13
    • one year ago
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    thank you both so much!!!! :)))

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