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  ## emcrazy14 3 years ago 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.
  
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  1. myininaya
  
  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. emcrazy14
  
  $y={a(x-a)^2}\div x^2-4a^2$
  
  3. emcrazy14
  
  I couldnot solve y'=0. I TRIED! :(
  
  4. myininaya
  
  What did you get for y'
  
  5. myininaya
  
  Oh and we also need to find where y' does not exist
  
  6. myininaya
  
  Use quotient rule.
  
  7. emcrazy14
  
  a(x-a)^2(2x-8a)-(x^2-4a^2)(2ax-2a^2)=0
  
  8. myininaya
  
  hmmm.... You know a is a constant right?
  
  9. emcrazy14
  
  Yep!
  
  10. myininaya
  
  Derivative of a is 0. Derivative of any constant *a is 0.
  
  11. myininaya
  
  a^2 is also a constant since a is a constant (a^2)'=0
  
  12. emcrazy14
  
  Oops. Now I get it. O_o
  
  13. myininaya
  
  And don't forget the quotient rules is : $y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2}$
  
  14. emcrazy14
  
  Can you please show me what exactly y' would be? All these 'a' and brackets have confused me. :(
  
  15. emcrazy14
  
  @experimentX : Can you please help me with this question?
  
  16. myininaya
  
  $y=\frac{a(x-a)^2}{x^2-4a^2}$ $\text{ What is } [a(x-a)^2]' \text{ equal to? }$ $\text{ What is } [x^2-4a^2]' \text{ equal to?}$
  
  17. myininaya
  
  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.
  
  18. myininaya
  
  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.
  
  19. myininaya
  
  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: $y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2}$
  
  20. myininaya
  
  So so far we have: $y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}$
  
  21. myininaya
  
  I'm just asking you now to give me the derivative of $(a(x-a))^2 \text{ and } (x^2-4a^2)$.
  
  22. emcrazy14
  
  For (a(x-a))^2, it would be 2x-2a. For (x^2-4a^2), it would be 2x.
  
  23. myininaya
  
  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?
  
  24. emcrazy14
  
  Would it be 2a(x-a)? :O
  
  25. myininaya
  
  Yep so we have $y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}$ = $y'=\frac{ ( 2a(x-a) ) \cdot (x^2-4a^2) - ( 2x ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}$ = $=\frac{2a(x-a)(x^2-4a^2)-2xa(x-a)^2}{(x^2-4a^2)^2}$
  
  26. myininaya
  
  Now what factors in the numerator do the two terms on top have in common?
  
  27. emcrazy14
  
  2a and x-a.
  
  28. myininaya
  
  Yes. :) So factor that out.
  
  29. myininaya
  
  $2a(x-a)(x^2-4a^2)-2xa(x-a)^2$ = $2a(x-a)[ ? - ?]$
  
  30. emcrazy14
  
  2a(x-a){(x^2-4a^2)-x(x-a)}
  
  31. myininaya
  
  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.
  
  32. myininaya
  
  $\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}$ You have like terms. :)
  
  33. myininaya
  
  Do you see the like terms to combine?
  
  34. emcrazy14
  
  Yep. I'll end up with 2a(x-a)(xa-4a^2). Right?
  
  35. myininaya
  
  Ok great now set equal factor equal to zero You are almost there :)
  
  36. myininaya
  
  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 :)
  
  37. emcrazy14
  
  So I'll have x=a and x=4a? What do I do now?
  
  38. myininaya
  
  It said find the coordinates so find what y is if x=a and find what y is if x=4a
  
  39. emcrazy14
  
  If x=a, then y=0. If x=4a, y=9a^3. Is it correct?
  
  40. emcrazy14
  
  No. Sorry. For x=4a, y is undefined.
  
  41. myininaya
  
  Your first one is right. Your second one is wrong.
  
  42. emcrazy14
  
  So y is not undefined?
  
  43. emcrazy14
  
  Oops. I found out my mistake. For x=4a, y=3a/4. Thank you soooooo muchhhhh @myininaya ! :D
  
  44. myininaya
  
  Great. :)
  
  45. emcrazy14
  
  One more thing, how do I show which is the minimum point and which is maximum? Is it by the second derivative?
  
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