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mathcalculus Group Title

help help help. Determine the extrema of f(x)= (-4)* x/ x^2+7 below on the given interval (a) on [1,4] The minimum is ?? and the maximum is ?? (b) on [1,5] The minimum is ?? and the maximum is ??

  • one year ago
  • one year ago

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  1. mathcalculus Group Title
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    do we use the quadratic formula for this one?

    • one year ago
  2. mattt9 Group Title
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    did you find the derivative?

    • one year ago
  3. mathcalculus Group Title
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    derivative is 1*x^2+7-2x*(-4)*x/ (x^2+7)^2

    • one year ago
  4. mathcalculus Group Title
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    from there.. x^2-6x+7/x^4+14x^2+49

    • one year ago
  5. mathcalculus Group Title
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    then i = to 0

    • one year ago
  6. mathcalculus Group Title
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    and i'm left with x^2-6x+7

    • one year ago
  7. mathcalculus Group Title
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    however, i can't factor this... so do i use quadratic formula?

    • one year ago
  8. mattt9 Group Title
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    yeah I did not check your math, however if it is correct then use the quadratic formula.

    • one year ago
  9. mathcalculus Group Title
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    can you try the problem also?

    • one year ago
  10. mathcalculus Group Title
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    i want to know if its correct.

    • one year ago
  11. zepdrix Group Title
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    "derivative is 1*x^2+7-2x*(-4)*x/ (x^2+7)^2" That first term shouldn't be a 1. The derivative of -4x is not 1. I think that should fix it up for you.

    • one year ago
  12. mathcalculus Group Title
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    yes i must of typed it wrong. but that's exactly what i got here.

    • one year ago
  13. zepdrix Group Title
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    \[\large f'(x)=\frac{\color{royalblue}{(-4)}(x^2+7)-(-4x)\color{royalblue}{(2x)}}{(x^2+7)^2}\]

    • one year ago
  14. mathcalculus Group Title
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    then after..

    • one year ago
  15. zepdrix Group Title
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    Set it equal to zero, then multiply both sides by the denominator. Don't do long division or anything silly like that.

    • one year ago
  16. mathcalculus Group Title
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    yeah i ended up with x^2-6x+7/x^4+14x^2+49

    • one year ago
  17. mathcalculus Group Title
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    ??

    • one year ago
  18. zepdrix Group Title
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    Sorry website wasn't working...

    • one year ago
  19. zepdrix Group Title
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    \[\large 0=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]Multiplying both sides by the denominator gives us,\[\large 0=-4x^2-28+8x^2\] I don't understand how you got the x term in the middle.

    • one year ago
  20. mathcalculus Group Title
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    how do i get the derivative. can you show me?

    • one year ago
  21. zepdrix Group Title
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    \[\large f(x)=\frac{-4x}{x^2+7}\] Remember the quotient rule, it will tell us to setup the derivative like this,\[\large f'(x)=\frac{\color{royalblue}{(-4x)'}(x^2+7)-(-4x)\color{royalblue}{(x^2+7)'}}{(x^2+7)^2}\]The blue terms are the ones we need to differentiate.

    • one year ago
  22. mathcalculus Group Title
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    yeah

    • one year ago
  23. mathcalculus Group Title
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    according to the quotient rule.

    • one year ago
  24. zepdrix Group Title
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    Which gives us this,\[\large f'(x)=\frac{\color{royalblue}{(-4)}(x^2+7)-(-4x)\color{royalblue}{(2x)}}{(x^2+7)^2}\] Which simplifies to this,\[\large f'(x)=\frac{-4x^2-28+8x^2}{(x^2+7)^2}\]Right?

    • one year ago
  25. mathcalculus Group Title
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    oh ok thank u!

    • one year ago
  26. mathcalculus Group Title
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    then it's 4x^2-28 right?

    • one year ago
  27. zepdrix Group Title
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    Yes. From there you can find a critical point.

    • one year ago
  28. mathcalculus Group Title
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    radical + or -7

    • one year ago
  29. mathcalculus Group Title
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    ?

    • one year ago
  30. mathcalculus Group Title
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    @zepdrix

    • one year ago
  31. mathcalculus Group Title
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    i appreciate your help a lot. i just need time to figure this out... :(

    • one year ago
  32. zepdrix Group Title
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    Yah that sounds right. So we have a couple steps now. We plug our `critical points` into the original function, and write down the \(\large f\) values they produce. Then, plug the `end points` into the original function, and write down the \(\large f\) values they produce. Then simply compare the \(\large f\) values. The largest will be your maximum. The smallest, your minimum.

    • one year ago
  33. zepdrix Group Title
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    \[\large x \in\left[1,4\right]\]These are our end points, 1 and 4. The end points of our interval.

    • one year ago
  34. mathcalculus Group Title
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    is the first critical point: -2 radical 7/7 ?

    • one year ago
  35. mathcalculus Group Title
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    thats the positive criticla point that i used.

    • one year ago
  36. mathcalculus Group Title
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    @zepdrix

    • one year ago
  37. zepdrix Group Title
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    yah that sounds right.

    • one year ago
  38. mathcalculus Group Title
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    and the f (- radical 7) is 4 radical 7?

    • one year ago
  39. mathcalculus Group Title
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    hope it's right :(

    • one year ago
  40. mathcalculus Group Title
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    are you checking to see if i am doing the problem correct?

    • one year ago
  41. mathcalculus Group Title
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    very worried, please help me.

    • one year ago
  42. zepdrix Group Title
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    no the negative root should produce 2sqrt7/7 i think.

    • one year ago
  43. mathcalculus Group Title
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    ?

    • one year ago
  44. mathcalculus Group Title
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    it's not \[\frac{ -2\sqrt{7} }{ 7 }\] ??

    • one year ago
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