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  1. #1
    But it was only fantasy SCAR-H's Avatar
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    Math experts of the misc GTFIH!!! (4k reps)

    Currently studying for my Calculus exam on Monday but am stuck on some particular problem.



    I took the first derivative and got:

    12x^2 + 30x + 18

    Then I factored it:

    6(2x^2 + 5x + 3)

    Used the quadratic formula and got x=-1 and x=-1.5

    My question is... what do I do after that? srs

    Thanks in advance.

    (I know there's an official math thread but I would literally have to wait a few days before getting an answer)
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  2. #2
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    you know the x-values of the extrema so plug them back into the original function and get your max/min values and see which are absolute
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    Plug your critical numbers into the original equation.

    First find f'(0).

    I didn't check your work but that would make -1 and -1.5 your critical numbers.

    f(-1)
    f(-1.5)
    f(0)
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    f'(x) = -12x^2 + 30x+18
    zeros = -.5,3
    plug em back into f(x)
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    The next step is to test the extremes you got by setting the derivative to = 0.

    just plug in a slightly larger/smaller value for x and see what you get in the function. this way you know if they are max's or mins
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    Originally Posted by cobhc View Post
    Plug your critical numbers into the original equation.

    First find f'(0).

    I didn't check your work but that would make -1 and -1.5 your critical numbers.

    f(-1)
    f(-1.5)
    f(0)
    Also check your boundary conditions.

    and FYI, the derivative is -12x^2 + 30x + 18

    Double check your answers by plugging it into your calculator and using the max,min thing (TI83)
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  7. #7
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    use critical numbers including -2 and 5 and plug them back into f(x) not f'(x)
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    But it was only fantasy SCAR-H's Avatar
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    Originally Posted by cobhc View Post
    Plug your critical numbers into the original equation.

    First find f'(0).

    I didn't check your work but that would make -1 and -1.5 your critical numbers.

    f(-1)
    f(-1.5)
    f(0)
    I did EXACTLY this, but the answer key from my professor shows completely different numbers... this is why I came here.

    This is what he wrote down.



    Am I missing a step after that?

    EDIT: Wtf how are you guys getting 5 and 3? :[
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  9. #9
    Founder of ESWKEAMSEC StrongScience62's Avatar
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    set the derivative equal to 0. solve for x. plug x back into original equation.

    edit: your derivative is completely wrong. da fug mayne.

    18 + 30x - 12x^2
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  10. #10
    Hamiltonian operator 4ea's Avatar
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    Originally Posted by SCAR-H View Post
    Currently studying for my Calculus exam on Monday but am stuck on some particular problem.



    I took the first derivative and got:

    12x^2 + 30x + 18

    Then I factored it:

    6(2x^2 + 5x + 3)

    Used the quadratic formula and got x=-1 and x=-1.5

    My question is... what do I do after that? srs

    Thanks in advance.

    (I know there's an official math thread but I would literally have to wait a few days before getting an answer)
    to distinguish between max and min take 2nd derivative. plug in your calculated value. if 2nd derivative is negative, its a max. if 2nd derivative is positive, its a min
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  11. #11
    Structural Brah cobhc's Avatar
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    Originally Posted by SCAR-H View Post
    I did EXACTLY this, but the answer key from my professor shows completely different numbers... this is why I came here.

    This is what he wrote down.



    Am I missing a step after that?
    Well for one, the roots of f'(x) are x = 3 and x = -.5
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  12. #12
    But it was only fantasy SCAR-H's Avatar
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    Originally Posted by StrongScience62 View Post
    set the derivative equal to 0. solve for x. plug x back into original equation.

    edit: your derivative is completely wrong. da fug mayne.

    18 + 30x - 12x^2
    Not sure if srs, you just typed the same thing. But thanks

    EDIT: Oh, I see that I ignored the negative sign on the original equation when I rearranged it.


    Originally Posted by 4ea View Post
    to distinguish between max and min take 2nd derivative. plug in your calculated value. if 2nd derivative is negative, its a max. if 2nd derivative is positive, its a min
    Alright, I will do that.
    Last edited by SCAR-H; 05-03-2012 at 11:15 PM.
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  13. #13
    Registered User gamblingman's Avatar
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    1. your first derivative is wrong, its f'(x)=-12x^2+30x-18

    which factors to -6(x-3)(2x+1)

    so critical numbers are, 3, -1/2 and the interval they gave u .

    so plug 3,-/2 and the end points of the interval into f(x)
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  14. #14
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    Originally Posted by moneymango View Post
    f'(x) = -12x^2 + 30x+18
    zeros = -.5,3
    plug em back into f(x)
    this is what i got. just plug -1/2 and 3 into the original equation.
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  15. #15
    Founder of ESWKEAMSEC StrongScience62's Avatar
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    Originally Posted by SCAR-H View Post
    Not sure if srs, you just typed the same thing. But thanks
    hey retard. negative signs matter. check again.
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  16. #16
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    Originally Posted by SCAR-H View Post
    Alright, I will do that.
    cool. i think you got the process down for getting the critical values. the second step with the 2nd derivatives is the final step for this problem
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  17. #17
    Structural Brah cobhc's Avatar
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    Also, don't forget that your intervals are critical numbers.

    f(-2)
    f(5)
    f(0)
    f(3)
    f(.5)

    And voila.
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  18. #18
    But it was only fantasy SCAR-H's Avatar
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    Originally Posted by gamblingman View Post
    1. your first derivative is wrong, its f'(x)=-12x^2+30x-18

    which factors to -6(x-3)(2x+1)

    so critical numbers are, 3, -1/2 and the interval they gave u .

    so plug 3,-/2 and the end points of the interval into f(x)
    The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol

    What do I look out for when factoring equations like these?
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    Originally Posted by SCAR-H View Post
    The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol

    What do I look out for when factoring equations like these?
    quadratic formula will always work. id say glance at the expression and see if the factoring jumps out at you, if it doesnt, just grind it out with quadratic formula
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    Structural Brah cobhc's Avatar
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    Originally Posted by SCAR-H View Post
    The main problem with these questions is that I suck ass at factoring. I tried using the quadratic formula... lol

    What do I look out for when factoring equations like these?
    Well your derivative before you simplify will be: -12x^2 + 30x + 18

    Pull a 6 out to make life easier.

    6(-2x^2 + 30x + 18)

    Factor out further.

    6(2x + 1)(x-3)

    2x + 1 = 0

    x = -1/2

    ----------------------------------

    x-3 = 0

    x = 3
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  21. #21
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    Originally Posted by cobhc View Post
    Well your derivative before you simplify will be: -12x^2 + 30x + 18

    Pull a 6 out to make life easier.

    6(-2x^2 + 30x + 18)

    Factor out further.

    6(2x + 1)(x-3)

    2x + 1 = 0

    x = -1/2

    ----------------------------------

    x-3 = 0

    x = 3
    It looks so simple once you have it all figured out. That's the only thing that keeps me going. Thanks

    EDIT: I have a question from a more recent test. I'm going to try it out and post it here to see if I get it right. Won't look at the answer key just yet.
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    Structural Brah cobhc's Avatar
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    Originally Posted by SCAR-H View Post
    It looks so simple once you have it all figured out. That's the only thing that keeps me going. Thanks
    Remember the formula: ax^2 + bx + c = (dx + e)(fx + g)
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    But it was only fantasy SCAR-H's Avatar
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    Alright, here we go.

    This is from the '11 test.



    f'(x)= 6x^2 - 18x - 24

    6(x^2 - 3x - 4)

    Quadratic formula

    x=4;x=-1

    x=4
    x=-1
    x=-2
    x=7

    Plug all of these numbers into the original equation and I get

    x@-2= 11
    x@-1= 28
    x@4= -97
    x@7= 92

    Abs max = f(7)
    Abs min = f(4)

    Does that look correct?
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    f'(x) = 6x^2 - 18x - 24

    6(x^2 -3x -4)

    6(x-4)(x+1)

    x=4
    x=-1

    f(4)
    f(-1)
    f(0)
    f(-2)
    f(7)

    I didn't plug them in but I think you're good to go OP. Good job.

    Just a something to remember for the exam, incase your professor tries to phuck with you:

    If any of your roots are not in the intervals given, then you don't have to plug them in.

    So if you had an x = 8 in this case, you would just ignore it.
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    Originally Posted by cobhc View Post
    f'(x) = 6x^2 - 18x - 24

    6(x^2 -3x -4)

    6(x-4)(x+1)

    x=4
    x=-1

    f(4)
    f(-1)
    f(0)
    f(-2)
    f(7)

    I didn't plug them in but I think you're good to go OP. Good job.

    Just a something to remember for the exam, incase your professor tries to phuck with you:

    If any of your roots are not in the intervals given, then you don't have to plug them in.

    So if you had an x = 8 in this case, you would just ignore it.
    Far from it, but thanks! It's covering all Chapter 4.

    L'Hospital
    Curve Sketching
    Optimization

    and a few other things.

    I'll do optimization tomorrow, since curve sketching seems a lot easier now.
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    Originally Posted by SCAR-H View Post
    Far from it, but thanks! It's covering all Chapter 4.

    L'Hospital
    Curve Sketching
    Optimization

    and a few other things.

    I'll do optimization tomorrow, since curve sketching seems a lot easier now.
    Feel free to pm with questions brah
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