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mathcalculus

  • one year ago

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 ??

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  1. mathcalculus
    • one year ago
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    do we use the quadratic formula for this one?

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

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

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

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

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

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

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

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

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

  11. zepdrix
    • one year ago
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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.

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

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

  14. mathcalculus
    • one year ago
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    then after..

  15. zepdrix
    • one year ago
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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.

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

  17. mathcalculus
    • one year ago
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    ??

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

  19. zepdrix
    • one year ago
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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.

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

  21. zepdrix
    • one year ago
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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.

  22. mathcalculus
    • one year ago
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    yeah

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

  24. zepdrix
    • one year ago
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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?

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

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

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

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

  29. mathcalculus
    • one year ago
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    ?

  30. mathcalculus
    • one year ago
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    @zepdrix

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

  32. zepdrix
    • one year ago
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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.

  33. zepdrix
    • one year ago
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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.

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

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

  36. mathcalculus
    • one year ago
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    @zepdrix

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

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

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

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

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

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

  43. mathcalculus
    • one year ago
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    ?

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

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