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emcrazy14 Group Title

PLEASE HELP! :( The curve C has equation y={a(x-a)^2}/x^2-4a^ where a is a positive constant. Show that C has one maximum point and one minimum point and find their coordinates.

  • 2 years ago
  • 2 years ago

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  1. myininaya Group Title
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    Find y' Set y' equal to 0 Solve y'=0 for x to get the critical numbers. Then we will determine if the critical numbers are max or mins or neither.

    • 2 years ago
  2. emcrazy14 Group Title
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    \[y={a(x-a)^2}\div x^2-4a^2\]

    • 2 years ago
  3. emcrazy14 Group Title
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    I couldnot solve y'=0. I TRIED! :(

    • 2 years ago
  4. myininaya Group Title
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    What did you get for y'

    • 2 years ago
  5. myininaya Group Title
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    Oh and we also need to find where y' does not exist

    • 2 years ago
  6. myininaya Group Title
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    Use quotient rule.

    • 2 years ago
  7. emcrazy14 Group Title
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    a(x-a)^2(2x-8a)-(x^2-4a^2)(2ax-2a^2)=0

    • 2 years ago
  8. myininaya Group Title
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    hmmm.... You know a is a constant right?

    • 2 years ago
  9. emcrazy14 Group Title
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    Yep!

    • 2 years ago
  10. myininaya Group Title
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    Derivative of a is 0. Derivative of any constant *a is 0.

    • 2 years ago
  11. myininaya Group Title
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    a^2 is also a constant since a is a constant (a^2)'=0

    • 2 years ago
  12. emcrazy14 Group Title
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    Oops. Now I get it. O_o

    • 2 years ago
  13. myininaya Group Title
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    And don't forget the quotient rules is : \[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2}\]

    • 2 years ago
  14. emcrazy14 Group Title
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    Can you please show me what exactly y' would be? All these 'a' and brackets have confused me. :(

    • 2 years ago
  15. emcrazy14 Group Title
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    @experimentX : Can you please help me with this question?

    • 2 years ago
  16. myininaya Group Title
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    \[y=\frac{a(x-a)^2}{x^2-4a^2}\] \[ \text{ What is } [a(x-a)^2]' \text{ equal to? }\] \[ \text{ What is } [x^2-4a^2]' \text{ equal to?}\]

    • 2 years ago
  17. myininaya Group Title
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    For my first question, a is a constant multiple. We can just bring that down and look at differentiating (x-a)^2 by using chain rule.

    • 2 years ago
  18. myininaya Group Title
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    For my second question, 4a^2 is just a constant so we only need to look at differentiating x^2 by use of the power rule.

    • 2 years ago
  19. myininaya Group Title
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    When we get done, we will plug in the bottom, the top, the derivative of the bottom, and the derivative of the top into the following formula I gave you earlier: \[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2} \]

    • 2 years ago
  20. myininaya Group Title
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    So so far we have: \[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \]

    • 2 years ago
  21. myininaya Group Title
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    I'm just asking you now to give me the derivative of \[ (a(x-a))^2 \text{ and } (x^2-4a^2) \].

    • 2 years ago
  22. emcrazy14 Group Title
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    For (a(x-a))^2, it would be 2x-2a. For (x^2-4a^2), it would be 2x.

    • 2 years ago
  23. myininaya Group Title
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    For the first one, couldn't you just leave it as 2(x-a) but don't forget to bring down that a in front. :) So you would actually have what for the first one?

    • 2 years ago
  24. emcrazy14 Group Title
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    Would it be 2a(x-a)? :O

    • 2 years ago
  25. myininaya Group Title
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    Yep so we have \[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \] = \[y'=\frac{ ( 2a(x-a) ) \cdot (x^2-4a^2) - ( 2x ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}\] = \[=\frac{2a(x-a)(x^2-4a^2)-2xa(x-a)^2}{(x^2-4a^2)^2}\]

    • 2 years ago
  26. myininaya Group Title
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    Now what factors in the numerator do the two terms on top have in common?

    • 2 years ago
  27. emcrazy14 Group Title
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    2a and x-a.

    • 2 years ago
  28. myininaya Group Title
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    Yes. :) So factor that out.

    • 2 years ago
  29. myininaya Group Title
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    \[2a(x-a)(x^2-4a^2)-2xa(x-a)^2\] = \[2a(x-a)[ ? - ?]\]

    • 2 years ago
  30. emcrazy14 Group Title
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    2a(x-a){(x^2-4a^2)-x(x-a)}

    • 2 years ago
  31. myininaya Group Title
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    Yep You will have \[\frac{2a(x-a)[(x^2-4a^2)-x(x-a)]}{(x^2-4a^2)^2}\] You can simplify what you have in brackets above though.

    • 2 years ago
  32. myininaya Group Title
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    \[\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}\] You have like terms. :)

    • 2 years ago
  33. myininaya Group Title
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    Do you see the like terms to combine?

    • 2 years ago
  34. emcrazy14 Group Title
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    Yep. I'll end up with 2a(x-a)(xa-4a^2). Right?

    • 2 years ago
  35. myininaya Group Title
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    Ok great now set equal factor equal to zero You are almost there :)

    • 2 years ago
  36. myininaya Group Title
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    a is a positive constant so 2a can never be zero so set x-a equal to zero and set xa-4a^2=0 Solve both for x to find the critical numbers :)

    • 2 years ago
  37. emcrazy14 Group Title
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    So I'll have x=a and x=4a? What do I do now?

    • 2 years ago
  38. myininaya Group Title
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    It said find the coordinates so find what y is if x=a and find what y is if x=4a

    • 2 years ago
  39. emcrazy14 Group Title
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    If x=a, then y=0. If x=4a, y=9a^3. Is it correct?

    • 2 years ago
  40. emcrazy14 Group Title
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    No. Sorry. For x=4a, y is undefined.

    • 2 years ago
  41. myininaya Group Title
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    Your first one is right. Your second one is wrong.

    • 2 years ago
  42. emcrazy14 Group Title
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    So y is not undefined?

    • 2 years ago
  43. emcrazy14 Group Title
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    Oops. I found out my mistake. For x=4a, y=3a/4. Thank you soooooo muchhhhh @myininaya ! :D

    • 2 years ago
  44. myininaya Group Title
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    Great. :)

    • 2 years ago
  45. emcrazy14 Group Title
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    One more thing, how do I show which is the minimum point and which is maximum? Is it by the second derivative?

    • 2 years ago
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