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Math Help - Graphing Question With Constants A,K

  1. #1
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    Graphing Question With Constants A,K

    I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

    y = x^k/1+x^k

    k is a positive integer greater than one. Here's what I get for the first derivative, after that I'm stuck:

    f'(x)= kx^(k-1)/(1+x^k)^2

    My guess is that the function is increasing for x>o and decreasing for x<0 but stuck after that.
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    Hello,

    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

    y = x^k/1+x^k

    k is a positive integer greater than one. Here's what I get for the first derivative, after that I'm stuck:

    f'(x)= kx^(k-1)/(1+x^k)^2

    My guess is that the function is increasing for x>o and decreasing for x<0 but stuck after that.
    This is not true...

    If k is odd, then k-1 is even \rightarrow k-1=2k'. Therefore, x^{k-1}=x^{2k'}=(x^{k'})^2, which is always positive !

    So you have to differentiate the possibilities "k is odd" & "k is even".
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  3. #3
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    I don't really understand. I'm differentiating with respect to x. So I wanna find the derivative for x^k to find where the function increases/decreases and then find the derivative of that to find where it's concave up/down.

    I thought the derivative of x^k would be kx^k-1. I dunno what the second derivative would be.
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I don't really understand. I'm differentiating with respect to x. So I wanna find the derivative for x^k to find where the function increases/decreases and then find the derivative of that to find where it's concave up/down.

    I thought the derivative of x^k would be kx^k-1. I dunno what the second derivative would be.
    But you can't know if x^{k-1} is actually positive or negative. It depends on both k and x.
    If x>0, there is no problem.

    If x<0, it will be positive if k-1 is positive. Otherwise, it will be negative.



    For the second derivative, it's the same thing. You have k \cdot x^{k-1}, which has k(k-1)x^{k-2} as derivative...
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  5. #5
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    Quote Originally Posted by SportfreundeKeaneKent View Post
    I have to find where y is increasing/decreasing, where it's concave up/down and any inflection points, if there are horizontal asymptotes, and then sketch it. Here's the problem:

    y = x^k/1+x^k
    Next time, please use parentheses/brackets or latex.

    I'm guessing you mean this:

    y=\frac{x^k}{1+x^k}

    So, using the quotient rule to find the derivative:

    y'=\frac{(\frac{d}{dx}x^k)(1+x^k)-[\frac{d}{dx}(1+x^k)](x^k)}{(1+x^k)^2}

    Derivate the terms on the right:

    y'=\frac{(kx^{k-1})(1+x^k)-kx^{k-1}(x^k)}{(1+x^k)^2}

    Simplify a little bit:

    y'=\frac{kx^{k-1}+kx^{k-1}(x^k)-kx^{k-1}(x^k)}{(1+x^k)^2}

    y'=\frac{kx^{k-1}}{(1+x^k)^2}

    Convert the form to make your life easier for the next derivative:

    y'=kx^{k-1}(1+x^k)^{-2}

    Now take the next derivative using the product rule:

    y''=kx^{k-1}[\frac{d}{dx}(1+x^k)^{-2}]+(1+x^k)^{-2}[\frac{d}{dx}kx^{k-1}]

    Derivate the first term using the chain rule:

    y''=kx^{k-1}(-2)(1+x^k)^{-3}kx^{k-1}+(1+x^k)^{-2}[\frac{d}{dx}kx^{k-1}]

    And then the second term:

    y''=kx^{k-1}(-2)(1+x^k)^{-3}kx^{k-1}+(1+x^k)^{-2}k(k-1)x^{k-2}

    Simplify a bit:

    y''=-2k^2x^{2k-2}(1+x^k)^{-3}+(1+x^k)^{-2}k(k-1)x^{k-2}

    That should get you started.
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