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Math Help - concavity...why?

  1. #1
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    concavity...why?

    I have a few problems that i don't understand the answer..

    f(x)=2x^2 -3x+4
    f'(x)= 4x-3
    f"(x)=4
    answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

    f(x)=x^4/7
    f'(x)= 4/7 x^-3/7
    f"(x)=-12/49 x^-10/7
    answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

    f(x)= root 4-x = (4-x)^1/2
    f'(x)=1/2 (4-x)^-1/2
    f"(x)=-1/4 (4-x)^-3/2
    amswer s cd (-infinity,4)......if i plug in say f(5) i'd get a positive..so why is it not cd(-infinity,4) AND cu (4, infinity)?

    thanks a bunch...i really appreciate u guys helping me, i only have bits and pieces of this stuff..but i'm getting the bigger picture a little at a time
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  2. #2
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    Quote Originally Posted by larasanut View Post
    I have a few problems that i don't understand the answer..

    f(x)=2x^2 -3x+4
    f'(x)= 4x-3
    f"(x)=4
    answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...

    Mr F says: Because the domain of f is all real numbers and f'' > 0 for all real numbers.

    f(x)=x^4/7
    f'(x)= 4/7 x^-3/7
    f"(x)=-12/49 x^-10/7
    answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)

    Mr F says: Because the domain of f is all real numbers and f'' < 0 for all real numbers except x = 0 (where it's undefined).

    f(x)= root 4-x = (4-x)^1/2
    f'(x)=1/2 (4-x)^-1/2
    f"(x)= -1/4 (4-x)^-3/2
    amswer s cd (-infinity,4)......if i plug in say f(5) i'd get a positive Mr F says: That's strange. Especially since x = 5 is not in the domain of f since f(5) is not defined. Concavity can obviously only exist where the fuction exists, that is, over the domain of the function.

    ..so why is it not cd(-infinity,4) AND cu (4, infinity)?

    thanks a bunch...i really appreciate u guys helping me, i only have bits and pieces of this stuff..but i'm getting the bigger picture a little at a time
    Your problems are a combination of carelessness and a lack of understanding of the defintion of concavity.
    Read this to address the latter problem: Concavity and Points of Inflection
    Last edited by mr fantastic; November 7th 2008 at 02:46 PM.
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  3. #3
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    Quote Originally Posted by larasanut View Post
    I have a few problems that i don't understand the answer..
    f(x)=2x^2 -3x+4
    f'(x)= 4x-3
    f"(x)=4
    answer is cu(-infinity, +infinity)...how do you determine that it is up? and why is it -infinity, infinity...
    You have found the derivatives correctly. So we know that if f(x) on [a,b] this implies that f''(x)>0\quad\forall{x}\in[a,b]

    So what interval is f(x)=4>0? It is obviously (-\infty,\infty). Or in other words every real number.

    f(x)=x^4/7
    f'(x)= 4/7 x^-3/7
    f"(x)=-12/49 x^-10/7
    answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)
    Consider this x^{\frac{-10}{7}}=\frac{1}{x^{\frac{10}{7}}}=\left(\frac{1}{  x^{\frac{5}{7}}}\right)^2

    So we know that this will be greater than zero forall values of x. So ifwe multiply it by a negative number it will be less than zero for all x.

    f(x)= \sqrt{4-x} = (4-x)^{\frac{1}{2}}
    f'(x)=\frac{1}{2} (4-x)^{\frac{-1}{2}}
    f"(x)=\frac{-1}{4} (4-x)^{\frac{-3}{2}}
    amswer s cd (-\infty,4)\......if i plug in say f(5) i'd get a positive..so why is it not cd  (-\infty,4) AND cu (4, infinity)?
    Your analysis would be correct except the values greater than four are imaginary.
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  4. #4
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    Mathstud..you helped me out tremendously on the first and last...but they second one i'm still confused on...i get its not defined at 0..but i still don't get why its concaved down...
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  5. #5
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    Quote Originally Posted by larasanut View Post
    I[snip]
    f(x)=x^4/7

    f'(x)= 4/7 x^-3/7

    f"(x)=-12/49 x^-10/7

    answer is cd(-infinity,0) U (0, infinity).....how do u plug numbers into f"(x) to determine which direction its concave... if i plug in f(-1) would it not be -12/49 (-1)^-10/7..but it would be positive...so how do they get (-)
    [snip]
    Quote Originally Posted by larasanut View Post
    Mathstud..you helped me out tremendously on the first and last...but they second one i'm still confused on...i get its not defined at 0..but i still don't get why its concaved down...
    Try reading my earlier reply (especially the bit about carelessness)!

    Note that (-1)^{-10/7} = \frac{1}{(-1)^{10/7}} = \frac{1}{[(-1)^{1/7}]^{10}} = (-1)^{10} = 1. Therefore f''(-1) = -\frac{12}{49} \cdot (1) < 0.

    You need to be much more careful with basic arithmetic.
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  6. #6
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    i get that if it was 1/(-1)^10/9...but i don't see how that is the same as -12/49(-1)^10/9...neg times a neg is pos...so why is it not upwarad
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  7. #7
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    i got...dur...
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  8. #8
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    i mean i got it so if it was say -12/49(-1)^-17/9 it would be upward
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  9. #9
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    Quote Originally Posted by larasanut View Post
    i get that if it was 1/(-1)^10/9...but i don't see how that is the same as -12/49(-1)^10/9...neg times a neg is pos...so why is it not upwarad
    *Sigh*

    Quote Originally Posted by mr fantastic View Post
    [snip]
    Note that (-1)^{-10/7} = \frac{1}{(-1)^{10/7}} = \frac{1}{[(-1)^{1/7}]^{10}} = (-1)^{10} = {\color{red}1}.
    [snip]
    Note very carefully. It's 1. NOT -1. When a negative number is raised to an even power you get a positive number. Not a negative number.

    So you have neg times pos = neg.

    It's all there in my previous reply if only you would take greater care in reading and processing it.
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