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    Hiya guys,
    I'm having trouble with a problem form my c3 text book. The question is:
    Show that the curve y=(x^3)/(1+x^4)^0.5 has a positive gradient for all values of x except at x=0
    I have differentiated the equation to find dy/dx however i cannot see how this will always be positive.
    Thanks
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    What's your dy/dx equation?
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    (Original post by noodlemon)
    Hiya guys,
    I'm having trouble with a problem form my c3 text book. The question is:
    Show that the curve y=(x^3)/(1+x^4)^0.5 has a positive gradient for all values of x except at x=0
    I have differentiated the equation to find dy/dx however i cannot see how this will always be positive.
    Thanks
    You should get

    \dfrac{dy}{dx} = \dfrac{x^2(x^4+3)}{(1+x^4)^{1.5}  }

    Now, notice that

    x^2 > 0

    for all non-zero x. So what can you say about x^4

    ?
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    originally, (3x^2(1+x^4)^0.5)-(2x^6(x^4+1))^0.5)/(x^4+1), which i simplified in the end to, x^2(3+x^4)/(x^4+1)^3/2 ?? i used the quotient rule.
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    (Original post by Indeterminate)
    You should get

    \dfrac{dy}{dx} = \dfrac{x^2(x^4+3)}{(1+x^4)^{1.5}  }

    Now, notice that

    x^2 > 0

    for all non-zero x. So what can you say about x^4

    ?
    yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?
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    (Original post by noodlemon)
    yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?
    Can 1 + x^4 be negative?
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    (Original post by ThatRandomGuy)
    Can 1 + x^4 be negative?
    nope, as x^4 is always positive, plus one therefore will be +ve too? however if you put that to the power of (3/2) im failing to realize why that's always +ve, ie, why is x^(3/2) always >0? as isnt that just square root (x^3), which surely is a plus or minus? im really confused here :s
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    (Original post by noodlemon)
    yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?
    What makes you think this? Do you see a plus-or-minus sign in front of the root?

    You only get positive and negative square roots when, for example, you use algebra to undo a squaring.
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    (Original post by noodlemon)
    yes i acknowledge the numerator to be always +ve, but the denominator, surely it can be -ve too as theres a root involved?
    Well, what goes wrong when you try to solve

    \dfrac{x^2(3+x^4)}{(1+x^4)^{1.5}  } < 0

    ?
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    (Original post by noodlemon)
    nope, as x^4 is always positive, plus one therefore will be +ve too? however if you put that to the power of (3/2) im failing to realize why that's always +ve, ie, why is x^(3/2) always >0? as isnt that just square root (x^3), which surely is a plus or minus? im really confused here :s
    If 1 + x^4 is always positive then it's cube will always be positive which means that it's square root will also always be positive.
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    (Original post by ThatRandomGuy)
    If 1 + x^4 is always positive then it's cube will always be positive which means that it's square root will also always be positive.
    ooh im sorry, im not deliberately trying to be dumb, or difficult. But say lets take (1+x^4)^3 to equal x, and we know for a fact now that x>0 , as stated before. Now we try to root this, surely a root of a +ve number, can be positive or negative? say we take x=9, surely root x is +-3?
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    (Original post by Indeterminate)
    Well, what goes wrong when you try to solve

    \dfrac{x^2(3+x^4)}{(1+x^4)^{1.5}  } < 0

    ?
    ah i see that, thank you, thats effectively proven it, but could you please read my lsat post which i quoted thatrandomguy. but thanks, i accept that
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    (Original post by noodlemon)
    ooh im sorry, im not deliberately trying to be dumb, or difficult. But say lets take (1+x^4)^3 to equal x, and we know for a fact now that x>0 , as stated before. Now we try to root this, surely a root of a +ve number, can be positive or negative? say we take x=9, surely root x is +-3?
    The square root of a number is definded as the positive root. If you raise a number to a power e.g. x^1.5 this can only be 1 number, not 2.
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    (Original post by noodlemon)
    ah i see that, thank you, thats effectively proven it, but could you please read my lsat post which i quoted thatrandomguy. but thanks, i accept that
    I already answered your question: the thing is that because you differentiated the square root sign here is a positive square root sign (take the positive solution only). The fact that two values x=±k satisfy x^2=k^2 is irrelevant here.
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    (Original post by noodlemon)
    ooh im sorry, im not deliberately trying to be dumb, or difficult. But say lets take (1+x^4)^3 to equal x, and we know for a fact now that x>0 , as stated before. Now we try to root this, surely a root of a +ve number, can be positive or negative? say we take x=9, surely root x is +-3?
    No, I think you are getting confused.

    If x^2 = 9 then x could equal +-3. However if x = 9 then root x will only be positive 3.
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    (Original post by noodlemon)
    ooh im sorry, im not deliberately trying to be dumb, or difficult. But say lets take (1+x^4)^3 to equal x, and we know for a fact now that x>0 , as stated before. Now we try to root this, surely a root of a +ve number, can be positive or negative? say we take x=9, surely root x is +-3?
    thatrandomguy's post was confusing tbh

    The square root of a number is defined as the +ve square root
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    (Original post by TenOfThem)
    thatrandomguy's post was confusing tbh

    The square root of a number is defined as the +ve square root
    Oh i see... wow ive never thought of it that way, sorry for slow reply, i was getting my head round it. So effectively ive defined it to be positive already :P
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    Thank you everyone for your help its much appreciated
 
 
 
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