anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • 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
  • anonymous
\[y={a(x-a)^2}\div x^2-4a^2\]
anonymous
  • anonymous
I couldnot solve y'=0. I TRIED! :(

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