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anonymous

  • 5 years ago

Application of Differentiation ( max and min values) Help me through these steps to find max and min.... f(x)= x/x^2+1, [0,2]

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  1. anonymous
    • 5 years ago
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    The max/min will occur at an end point of your interval or at an interior point where \[f^{\prime}(x)=0\]

  2. anonymous
    • 5 years ago
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    what is f(x) in this case?

  3. anonymous
    • 5 years ago
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    is it\[f(x)=\frac{x}{x^2+1}\]?

  4. anonymous
    • 5 years ago
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    it is difficult to tell by the way you wrote it...

  5. anonymous
    • 5 years ago
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    yes but also [0,2]

  6. anonymous
    • 5 years ago
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    yes, I get that your are looking for absolute max/min on the interval [0,2]

  7. anonymous
    • 5 years ago
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    okay so first we need \[f^\prime(x)\]

  8. anonymous
    • 5 years ago
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    do you have it ? or do we need to figure it out?

  9. anonymous
    • 5 years ago
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    \[f^\prime(x)=\frac{1-x^2}{(x^2+1)^2}\]

  10. anonymous
    • 5 years ago
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    yes its 1(x^2+10 * x(2x+1) / (x^2+1)^2 right....i dont know how to write it out how you are

  11. anonymous
    • 5 years ago
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    you can use the equation writer button or encode directly with Latex as I am, but its okay to use plain text as long as you use grouping symbols to include all of the numerator and denominator

  12. anonymous
    • 5 years ago
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    your derivative does not look correct

  13. anonymous
    • 5 years ago
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    \[f^\prime(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1)^2}\]

  14. anonymous
    • 5 years ago
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    i so knew that but rushing

  15. anonymous
    • 5 years ago
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    \[=\frac{x^2+1-2x^2}{(x^2+1)^2}\]

  16. anonymous
    • 5 years ago
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    okay, cool

  17. anonymous
    • 5 years ago
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    then we can move forward

  18. anonymous
    • 5 years ago
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    since the denominator is the sum of positive numbers the derivative of this function is never undefined so we just need to find where it equals 0

  19. anonymous
    • 5 years ago
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    setting the numerator equal to zero we get \[x=\pm 1\]

  20. NY,NY
    • 5 years ago
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    habachuchu

  21. anonymous
    • 5 years ago
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    2x^2 / x^2+1 is what i get next

  22. anonymous
    • 5 years ago
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    now note that \[x=-1\] is not in the interval [0,2] so we do not need to consider it for this problem

  23. anonymous
    • 5 years ago
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    now we just evaluate \[f(0),f(1),f(2)\] and compare

  24. anonymous
    • 5 years ago
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    \[f(0)=0\]\[f(1)=\frac{1}{2}\]\[f(2)=\frac{2}{5}\]

  25. anonymous
    • 5 years ago
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    so on the interval [0,2] our function has an absolute max of 1/2 and an absolute min of 0

  26. anonymous
    • 5 years ago
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    okay i get the last part you did but after i get the x^2+1-2x^2 / (x^2+1)^2 i get confused on what you mean

  27. anonymous
    • 5 years ago
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    the numerator of your derivative is 1-x^2 after you "clean it up"

  28. anonymous
    • 5 years ago
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    thats the thing where i really need help my college algebra suck i dont know how to get that after i get 2x^2 / x^2+1

  29. anonymous
    • 5 years ago
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    x^2+1-2x^2=1-x^2

  30. anonymous
    • 5 years ago
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    okay i can see that i just thought you would canceled the numerator x^2+1 with the denominator and the square would cancel

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