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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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The text passage is a conversation between a student and a tutor, discussing a calculus problem involving finding the maximum and minimum points of a curve. The student asks for help, and the tutor guides them through the process of finding the derivative, solving for critical points, and determining the coordinates of the maximum and minimum points.
no_question

the passage does not contain a mathematical question in the form of a problem to be solved by the reader. instead, it is a dialogue where a tutor is guiding a student through solving a problem. the problem itself is not posed as a question but rather as a task to be completed.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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Here's the question you clicked on:

## 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.

• This Question is Closed
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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The text passage is a conversation between a student and a tutor, discussing a calculus problem involving finding the maximum and minimum points of a curve. The student asks for help, and the tutor guides them through the process of finding the derivative, solving for critical points, and determining the coordinates of the maximum and minimum points.

TYPE: NO_QUESTION

The passage does not contain a mathematical question in the form of a problem to be solved by the reader. Instead, it is a dialogue where a tutor is guiding a student through solving a problem. The problem itself is not posed as a question but rather as a task to be completed.