## 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. one year ago one year ago

1. myininaya Group Title

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 Group Title

$y={a(x-a)^2}\div x^2-4a^2$

3. emcrazy14 Group Title

I couldnot solve y'=0. I TRIED! :(

4. myininaya Group Title

What did you get for y'

5. myininaya Group Title

Oh and we also need to find where y' does not exist

6. myininaya Group Title

Use quotient rule.

7. emcrazy14 Group Title

a(x-a)^2(2x-8a)-(x^2-4a^2)(2ax-2a^2)=0

8. myininaya Group Title

hmmm.... You know a is a constant right?

9. emcrazy14 Group Title

Yep!

10. myininaya Group Title

Derivative of a is 0. Derivative of any constant *a is 0.

11. myininaya Group Title

a^2 is also a constant since a is a constant (a^2)'=0

12. emcrazy14 Group Title

Oops. Now I get it. O_o

13. myininaya Group Title

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 Group Title

Can you please show me what exactly y' would be? All these 'a' and brackets have confused me. :(

15. emcrazy14 Group Title

@experimentX : Can you please help me with this question?

16. myininaya Group Title

$y=\frac{a(x-a)^2}{x^2-4a^2}$ $\text{ What is } [a(x-a)^2]' \text{ equal to? }$ $\text{ What is } [x^2-4a^2]' \text{ equal to?}$

17. myininaya Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

I'm just asking you now to give me the derivative of $(a(x-a))^2 \text{ and } (x^2-4a^2)$.

22. emcrazy14 Group Title

For (a(x-a))^2, it would be 2x-2a. For (x^2-4a^2), it would be 2x.

23. myininaya Group Title

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 Group Title

Would it be 2a(x-a)? :O

25. myininaya Group Title

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 Group Title

Now what factors in the numerator do the two terms on top have in common?

27. emcrazy14 Group Title

2a and x-a.

28. myininaya Group Title

Yes. :) So factor that out.

29. myininaya Group Title

$2a(x-a)(x^2-4a^2)-2xa(x-a)^2$ = $2a(x-a)[ ? - ?]$

30. emcrazy14 Group Title

2a(x-a){(x^2-4a^2)-x(x-a)}

31. myininaya Group Title

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 Group Title

$\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}$ You have like terms. :)

33. myininaya Group Title

Do you see the like terms to combine?

34. emcrazy14 Group Title

Yep. I'll end up with 2a(x-a)(xa-4a^2). Right?

35. myininaya Group Title

Ok great now set equal factor equal to zero You are almost there :)

36. myininaya Group Title

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 Group Title

So I'll have x=a and x=4a? What do I do now?

38. myininaya Group Title

It said find the coordinates so find what y is if x=a and find what y is if x=4a

39. emcrazy14 Group Title

If x=a, then y=0. If x=4a, y=9a^3. Is it correct?

40. emcrazy14 Group Title

No. Sorry. For x=4a, y is undefined.

41. myininaya Group Title

Your first one is right. Your second one is wrong.

42. emcrazy14 Group Title

So y is not undefined?

43. emcrazy14 Group Title

Oops. I found out my mistake. For x=4a, y=3a/4. Thank you soooooo muchhhhh @myininaya ! :D

44. myininaya Group Title

Great. :)

45. emcrazy14 Group Title

One more thing, how do I show which is the minimum point and which is maximum? Is it by the second derivative?