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Math Help - Critical Numbers

  1. #1
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    Critical Numbers

    Can someone please help me solve this?

    F(x)=2x^3 - 3x^2 +36x +7

    I need to find all critical numbers. Thank you
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dimanre View Post
    Can someone please help me solve this?

    F(x)=2x^3 - 3x^2 +36x +7

    I need to find all critical numbers. Thank you

    Set f'(x)=0

    The values of x that satisfy this equation are your "critical #'s"
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  3. #3
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    ok

    so then I take 2(0)^3 - 3(0)^2 +36(0) +7

    so that gives me 7? since all others become zero? or am i way off?
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  4. #4
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    Quote Originally Posted by dimanre View Post
    so then I take 2(0)^3 - 3(0)^2 +36(0) +7

    so that gives me 7? since all others become zero? or am i way off?
    You have to take the derivative first, then set it to zero, and then solve for x. That will you give your critical numbers.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dimanre View Post
    so then I take 2(0)^3 - 3(0)^2 +36(0) +7

    so that gives me 7? since all others become zero? or am i way off?
    f'(x)=6x^2-6x+36

    Do you know how I did this?
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  6. #6
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    ok

    ok so it becomes 6x^2 - 6x +36=0

    is this correct? if so..what is next step?
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  7. #7
    Member McScruffy's Avatar
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    Quote Originally Posted by dimanre View Post
    ok so it becomes 6x^2 - 6x +36=0

    is this correct? if so..what is next step?
    Solve for x.
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  8. #8
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    Ok..i figured out how to get the derivative...and then i know you have to set x to zero..so do you multiply each by zero? thereby giving you zero for each section?
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  9. #9
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dimanre View Post
    Ok..i figured out how to get the derivative...and then i know you have to set x to zero..so do you multiply each by zero? thereby giving you zero for each section?
    Not quite.

    Set f'(x)=0 and then solve for x. At this point it's algebra man.
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  10. #10
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    yes...which is very weak....

    I know...but my algebra is horribly bad and rusty..havent taken it in 18 years....so i am stuck trying to figure it out...
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  11. #11
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    I believe it is 6(x^2-x+6) is that the final answer?
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  12. #12
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dimanre View Post
    I know...but my algebra is horribly bad and rusty..havent taken it in 18 years....so i am stuck trying to figure it out...
    6x^2-6x+36=0

    x^2-x+6=0

    x=\frac{1\pm\sqrt{1-4(1)(6)}}{2} (by the quadratic formula).

    You can see that this has no solution, therefore the are no critical numbers.
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  13. #13
    No one in Particular VonNemo19's Avatar
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    You can clearly see from the graph that f(x) does not have a place where a tangent line will be horizontal.
    Last edited by VonNemo19; September 19th 2009 at 10:49 PM.
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  14. #14
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    THank you so much for your help..i really appreciate it
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  15. #15
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by dimanre View Post
    THank you so much for your help..i really appreciate it
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