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Thread: find all extremum points

  1. #1
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    find all extremum points

    $\displaystyle (a) \frac{x^2}{1 + x^4}$ over $\displaystyle (-\infty,\infty).$

    $\displaystyle (b) \frac{\ln(x)}{x}$ over $\displaystyle [1,\infty).$
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by cgiulz View Post
    $\displaystyle (a) \frac{x^2}{1 + x^4}$ over $\displaystyle (-\infty,\infty).$

    $\displaystyle (b) \frac{\ln(x)}{x}$ over $\displaystyle [1,\infty).$
    What are you having trouble with here? What have you tried? Is it taking the derivative? Choosing test values? etc...?
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  3. #3
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    Try them out and you'll see. For $\displaystyle (a)$ the criticals are 0,1 and -1, and as $\displaystyle x \rightarrow \infty, f(x) \rightarrow 0,$ so might this be a minimum? Also, +-1 seem to be a single max? Moreover, the first and second derivative test fail..etc
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  4. #4
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    I think I got $\displaystyle (a)$ -- 0 is a max and +-1 are neither.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by cgiulz View Post
    Try them out and you'll see. For $\displaystyle (a)$ the criticals are 0,1 and -1, and as $\displaystyle x \rightarrow \infty, f(x) \rightarrow 0,$ so might this be a minimum? Also, +-1 seem to be a single max? Moreover, the first and second derivative test fail..etc
    a) $\displaystyle f'(x)=\frac{2x(1+x^4)-4x^3(x^2)}{(1+x^4)^2}=0\Rightarrow{x(1-x)(1+x)(1+x^2)}=0\Rightarrow$ critical numbers at $\displaystyle x=1,-1,0$

    So, evaluate f prime within these intervals and determine its sign.
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