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.

- anonymous

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- 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.

- anonymous

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

- anonymous

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

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## More answers

- myininaya

What did you get for y'

- myininaya

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

- myininaya

Use quotient rule.

- anonymous

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

- myininaya

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

- anonymous

Yep!

- myininaya

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

- myininaya

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

- anonymous

Oops. Now I get it. O_o

- myininaya

And don't forget the quotient rules is :
\[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2}\]

- anonymous

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

- anonymous

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

- myininaya

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

- 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.

- 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.

- 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:
\[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2} \]

- myininaya

So so far we have:
\[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \]

- myininaya

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

- anonymous

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

- 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?

- anonymous

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

- myininaya

Yep so we have
\[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \]
=
\[y'=\frac{ ( 2a(x-a) ) \cdot (x^2-4a^2) - ( 2x ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}\]
=
\[=\frac{2a(x-a)(x^2-4a^2)-2xa(x-a)^2}{(x^2-4a^2)^2}\]

- myininaya

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

- anonymous

2a and x-a.

- myininaya

Yes. :) So factor that out.

- myininaya

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

- anonymous

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

- 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.

- myininaya

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

- myininaya

Do you see the like terms to combine?

- anonymous

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

- myininaya

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

- 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 :)

- anonymous

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

- myininaya

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

- anonymous

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

- anonymous

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

- myininaya

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

- anonymous

So y is not undefined?

- anonymous

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

- myininaya

Great. :)

- anonymous

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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