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mathcalculus

  • 2 years 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
    • 2 years ago
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    do we use the quadratic formula for this one?

  2. mattt9
    • 2 years ago
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    did you find the derivative?

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

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

  5. mathcalculus
    • 2 years ago
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    then i = to 0

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

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

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

  9. mathcalculus
    • 2 years ago
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    can you try the problem also?

  10. mathcalculus
    • 2 years ago
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    i want to know if its correct.

  11. zepdrix
    • 2 years 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
    • 2 years ago
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    yes i must of typed it wrong. but that's exactly what i got here.

  13. zepdrix
    • 2 years 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
    • 2 years ago
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    then after..

  15. zepdrix
    • 2 years 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
    • 2 years ago
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    yeah i ended up with x^2-6x+7/x^4+14x^2+49

  17. mathcalculus
    • 2 years ago
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    ??

  18. zepdrix
    • 2 years ago
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    Sorry website wasn't working...

  19. zepdrix
    • 2 years 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
    • 2 years ago
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    how do i get the derivative. can you show me?

  21. zepdrix
    • 2 years 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
    • 2 years ago
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    yeah

  23. mathcalculus
    • 2 years ago
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    according to the quotient rule.

  24. zepdrix
    • 2 years 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
    • 2 years ago
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    oh ok thank u!

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

  27. zepdrix
    • 2 years ago
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    Yes. From there you can find a critical point.

  28. mathcalculus
    • 2 years ago
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    radical + or -7

  29. mathcalculus
    • 2 years ago
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    ?

  30. mathcalculus
    • 2 years ago
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    @zepdrix

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

  32. zepdrix
    • 2 years 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
    • 2 years 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
    • 2 years ago
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    is the first critical point: -2 radical 7/7 ?

  35. mathcalculus
    • 2 years ago
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    thats the positive criticla point that i used.

  36. mathcalculus
    • 2 years ago
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    @zepdrix

  37. zepdrix
    • 2 years ago
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    yah that sounds right.

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

  39. mathcalculus
    • 2 years ago
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    hope it's right :(

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

  41. mathcalculus
    • 2 years ago
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    very worried, please help me.

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

  43. mathcalculus
    • 2 years ago
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    ?

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

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