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AP Style Problem - Derivatives

  1. Oct 8, 2007 #1
    Hey guys. These are the first AP Problems I'm doing. Here is all my work.

    Let f be the function given by f(x) = x[tex]^{3}[/tex]-7x+6.

    a. Find the zeros of f.
    b. Write an equation of the line tangent to the graph of f at x = -1
    c. Find the number c that satisfies the Mean Value Theorem for f on the closed interval [1,3]

    So what I did is:
    a. 0 = x[tex]^{3}[/tex] - 7x + 6
    -6 = x(x[tex]^{2}[/tex]-7)
    Final Answer: x = +/-1, 6

    b. f'(x) = 3x[tex]^{2}[/tex] - 7
    m = f'(-1) = 3(-1)[tex]^{3}[/tex] - 7 = -4
    f'(-1) = (-1)[tex]^{3}[/tex] - 7(-1) + 6 = 12
    Final Answer: y - 12 = -4(x+1)

    c. f(b) = f(3) = 3[tex]^{3}[/tex] - 7(3) + 6 = -6
    f(a) = f(1) = 1[tex]^{3}[/tex] - 7(1) + 6 = 0
    f'(c) = (f(b)-f(a)) / (b-a) = (-6 - 0) / (3-1) = -3
    -3 = 3c[tex]^{2}[/tex] - 7
    Final Answer: c = 2[tex]\sqrt{3}[/tex] / 3

    Please tell me what is right and wrong? I am fairly confident on a and b, but shaky on c because I haven't studied that yet and don't know if it's right at all.

    Thanks in advance!
     
  2. jcsd
  3. Oct 8, 2007 #2

    HallsofIvy

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    I don't understand where you would get +/- 1, 6 from "[itex]-6= x(x^2- 7)[/itex] or even how that last form helps. Even if you don't know how to solve this equation, you should be able to check: If x= 1, then [itex]1^3- 7(1)+ 6= 0[/itex] so that is a solution. If x= -1, [itex](-1)^3- 7(-1)+ 6= -2[/itex] not 0. Finally, if x= 6, [itex]6^3- 7(6)+ 6= 216- 42+ 6= 180[/itex], nowhere near 0!. Since 1 is a root, you can divide [itex]x^3- 7x+ 6[/itex] by x-1 and find that [itex]x^3- 7x+ 6= (x-1)(x^2+ x- 6)[/itex]. what are the other two roots?

    you mean [itex]3(-1)^2- 7= -4[/itex] but yes, that is the correct slope.

    Well, you mean f(-1), not f'(-1), but even then, [itex](-1)^3- 7(-1)+ 6= -1-7+ 6= -2[/itex] not 12!

    Try again!

    ?? [itex]3^3 -7(3)+ 6= 27- 21+ 6= 12[/itex]!

    Your calculus is good but your arithmetic is atrocious! Perhaps you are trying to do it too fast.
     
  4. Oct 8, 2007 #3
    Lol thanks, I guess. Yes, I was in a rush to leave. Thank you for your help. I will go back and check everything. Hectic day, my friend. Forgive me for the careless mistakes.
     
  5. Oct 9, 2007 #4
    So I went back and reworked it.
    For a) I did the synthetic division and got "x = 1,2, -3" for my final answer.

    For b) Don't want to be rude, but my answer was correct. Your mistake was in the underlined part (-1)^3 -7(-1)+ 6= -1-(-7)+ 6= 12. Subtracting a negative (when you just subtracted the positive) or how ever way you would like to look at it.

    c) Yes, my math was horrible there.
    f(b) = f(3) = 3[tex]^{3}[/tex] - 7(3) + 6 = 12
    f(a) = f(1) = 1[tex]^{3}[/tex] - 7(1) + 6 = 0
    f'(c) = (f(b)-f(a)) / (b-a) = (12 - 0) / (3-1) = 6
    6 = 3c[tex]^{2}[/tex] - 7
    c = [tex]\sqrt{13/3}[/tex]

    I'm quite sure it's fixed now! Thanks for all the help!
     
  6. Oct 10, 2007 #5

    HallsofIvy

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    It is never rude to point out a critical error!
     
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