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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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.

Mathematics
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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.
\[y={a(x-a)^2}\div x^2-4a^2\]
I couldnot solve y'=0. I TRIED! :(

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What did you get for y'
Oh and we also need to find where y' does not exist
Use quotient rule.
a(x-a)^2(2x-8a)-(x^2-4a^2)(2ax-2a^2)=0
hmmm.... You know a is a constant right?
Yep!
Derivative of a is 0. Derivative of any constant *a is 0.
a^2 is also a constant since a is a constant (a^2)'=0
Oops. Now I get it. O_o
And don't forget the quotient rules is : \[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2}\]
Can you please show me what exactly y' would be? All these 'a' and brackets have confused me. :(
@experimentX : Can you please help me with this question?
\[y=\frac{a(x-a)^2}{x^2-4a^2}\] \[ \text{ What is } [a(x-a)^2]' \text{ equal to? }\] \[ \text{ What is } [x^2-4a^2]' \text{ equal to?}\]
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.
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.
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: \[y'=\frac{ (\text{ derivative of top }) \cdot (\text{ bottom } ) - ( \text{ derivative of bottom } ) \cdot ( \text{ top } )}{(bottom)^2} \]
So so far we have: \[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \]
I'm just asking you now to give me the derivative of \[ (a(x-a))^2 \text{ and } (x^2-4a^2) \].
For (a(x-a))^2, it would be 2x-2a. For (x^2-4a^2), it would be 2x.
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?
Would it be 2a(x-a)? :O
Yep so we have \[y'=\frac{ (\text{ derivative of top }) \cdot (x^2-4a^2) - ( \text{ derivative of bottom } ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2} \] = \[y'=\frac{ ( 2a(x-a) ) \cdot (x^2-4a^2) - ( 2x ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}\] = \[=\frac{2a(x-a)(x^2-4a^2)-2xa(x-a)^2}{(x^2-4a^2)^2}\]
Now what factors in the numerator do the two terms on top have in common?
2a and x-a.
Yes. :) So factor that out.
\[2a(x-a)(x^2-4a^2)-2xa(x-a)^2\] = \[2a(x-a)[ ? - ?]\]
2a(x-a){(x^2-4a^2)-x(x-a)}
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.
\[\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}\] You have like terms. :)
Do you see the like terms to combine?
Yep. I'll end up with 2a(x-a)(xa-4a^2). Right?
Ok great now set equal factor equal to zero You are almost there :)
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 :)
So I'll have x=a and x=4a? What do I do now?
It said find the coordinates so find what y is if x=a and find what y is if x=4a
If x=a, then y=0. If x=4a, y=9a^3. Is it correct?
No. Sorry. For x=4a, y is undefined.
Your first one is right. Your second one is wrong.
So y is not undefined?
Oops. I found out my mistake. For x=4a, y=3a/4. Thank you soooooo muchhhhh @myininaya ! :D
Great. :)
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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