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Math Help - Max/Min

  1. #1
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    Max/Min

    Let f be the function defined by f(x) = ln absolute value of  \frac{x}{1+ x ^2}

    b) Determine whether f is an even function, odd function or neither. Justify.

    c) At what values does f have relative maximum or minimum? Use first derivative test to determine whether f(x) is a relative max/ min.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ny_chow View Post
    Let f be the function defined by f(x)=\left|\frac{x}{1+ x ^2}\right|

    b) Determine whether f is an even function, odd function or neither. Justify.

    c) At what values does f have relative maximum or minimum? Use first derivative test to determine whether f(x) is a relative max/ min.
    b) Well, 1+x^2>0 for all x, so f(x) = \frac{|x|}{1+x^2}

    To be even, f(x)=f(-x), which is true here, because |x| is even, 1+x^2 is even, and the composition of even functions is even.

    c) f'(x) = \frac{(x^2+1)\cdot sign(x)-|x|(2x)}{(1+x^2)^2} = \frac{(x^2+1)\cdot sign(x)-(2x^2)\cdot sign(x)}{(1+x^2)^2} = sign(x)\cdot\frac{x^2+1-2x^2}{(1+x^2)^2} = sign(x)\cdot\frac{1-x^2}{(1+x^2)^2}

    Solving f'(x)=0 gives x=\pm 1.

    Taking the second derivative (just use the quotient rule and treat sign(x) as a constant) gives you sign(x)\cdot\frac{2x(x^2-3)}{(1+x^2)^3}. Plugging in x=1 and x=-1 both give you negative values so they are both local maximums.

    It is also worth noting that the sign(0)=0 so f'(0)=0. This is a local minimum as zero is the unique point for which f(x)=0; at all other values f(x)>0.
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  3. #3
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    what does
    sign (x) mean?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by ny_chow View Post
    what does
    sign (x) mean?
    It just tells you the sign of the number. It's explicit formula is:
    <br />
sign(x)=\left\{\begin{array}{lr}1&:x>0\\0&:x=0\\-1&:x<0\end{array}\right\}
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