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Math Help - Last Critical Point Question

  1. #1
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    Last Critical Point Question

    This is the last question in my problem set.

    "For the function f(x) = 3x^4 + ax^3 + bx^2 + cx + d,

    a. find constants a, b, c, and d that guarantee that the graph of f will have horizontal tangents at (-2, -73) and (0, -9).

    b. there is a third point that has a horizontal tangent. Find this point.

    c. For all three points, determine whether each corresponds to a relative maximum, a relative minimum, or neither."

    My only problem starting this question is what to do with the information given in part a. If someone could provide me a hint, it would be much obliged.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Here is what you do

    f(x)=3x^4+ax^3+bx^2+cx+d...thus f'(x)=12x^3+3ax^2+2bx+c...now to have a critical point at (-2,-73) you must set the derivative equal to zero 12x^3+3ax^2+2bx+c=0...hint hint and then combine that with the fact that it must also be true that f(-2)=-73...hint hint..and you also have that f'(0)=0...big hint big hint...and that f(0)=9

    you haev four equations and four variables..hmm this seems solvable...good luck hope this is hint enough..hint hint
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    f(x)=3x^4+ax^3+bx^2+cx+d...thus f'(x)=12x^3+3ax^2+2bx+c...now to have a critical point at (-2,-73) you must set the derivative equal to zero 12x^3+3ax^2+2bx+c=0...hint hint and then combine that with the fact that it must also be true that f(-2)=-73...hint hint..and you also have that f'(0)=0...big hint big hint...and that f(0)=9

    you haev four equations and four variables..hmm this seems solvable...good luck hope this is hint enough..hint hint
    we also have f'(-2) = 0
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  4. #4
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    Okay, lot's of hints!

    I've gotten this far...

    f(0) = -9
    -9 = 3(0)^4 + a(0)^3 + b(0)^2 + c(0) + d
    -9 = d

    f'(0) = 12(0)^3 + 3a(0)^2 + 2b(0) + c
    0 = c

    I can't figure out how to continue. :|
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  5. #5
    MHF Contributor Mathstud28's Avatar
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    Do the same thing

    except use x=-2 instead
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    Thanks to both Jhevon and Mathstud28. I've finished a and b and am moving to c.
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Anytime

    Math is my love...and you are just giving me practice on things I've learned which is reinforcing it!...so really thank you...because if it wasn't for people like you who need math help...well...I'd be really really bored haha
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Thanks to both Jhevon and Mathstud28. I've finished a and b and am moving to c.
    um, you already found that c was 0, didn't you?
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  9. #9
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    Quote Originally Posted by Jhevon View Post
    um, you already found that c was 0, didn't you?
    Haha, I meant part c of the question, not the c value! :P

    By the way, I'm a bit stuck with part c as well. :|
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    To do this

    since all the points were points that made f'(x)=0 test them in f''(x) if the point is positive its a min if its negative its a max...if its neither use test intervals based on f''(x) to determine if its an inflection point...or just say its neither xD
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jeavus View Post
    Haha, I meant part c of the question, not the c value! :P

    By the way, I'm a bit stuck with part c as well. :|
    hehe, ok, my bad.

    there are several ways to attack part c, but it depends on what you've learned.

    you can use the second derivative, that is, find f''(x). plug in the x-values of you critical points in that equation. if f''(x) > 0, then you have a relative minimum, if f''(x) < 0, you have a relative maximum
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  12. #12
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    Thanks again, to both of you.
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  13. #13
    MHF Contributor Mathstud28's Avatar
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    Can I ask you a question

    I only ask question on here that are just out of curiosity..that came to be because of my mathematical tinkering...but you ask for Homework...don't teachers get weary that you are utilizing concepts that are currently beyond you?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    I only ask question on here that are just out of curiosity..that came to be because of my mathematical tinkering...but you ask for Homework...don't teachers get weary that you are utilizing concepts that are currently beyond you?
    some are not, but some want you to learn on your own. that's all well and good if they prepare you for it. but then there are teachers who all they do is teach you to follow methods, "don't worry about why this is true, just follow the formula", and then they turn around and expect you to be able to learn new concepts on your own. i think that's crazy. teach kids to think, then there's no harm in giving them material that is a bit beyond them
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  15. #15
    MHF Contributor Mathstud28's Avatar
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    Haha

    that sounds like my current geometry teacher haha
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