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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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! :(

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.