Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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

Answer 4

What did you get for y'

Answer 5

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

Answer 6

Use quotient rule.

Answer 7

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

Answer 8

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

Answer 9

Yep!

Answer 10

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

Answer 11

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

Answer 12

Oops. Now I get it. O_o

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

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

Answer 17

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.

Answer 18

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.

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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?

Answer 24

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

Answer 25

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

=

\[y'=\frac{ ( 2a(x-a) ) \cdot (x^2-4a^2) - ( 2x ) \cdot ( a(x-a)^2 )}{(x^2-4a^2)^2}\]

=

\[=\frac{2a(x-a)(x^2-4a^2)-2xa(x-a)^2}{(x^2-4a^2)^2}\]

Answer 26

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

Answer 27

2a and x-a.

Answer 28

Yes. :) So factor that out.

Answer 29

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

Answer 30

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

Answer 31

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.

Answer 32

\[\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}\]

You have like terms. :)

Answer 33

Do you see the like terms to combine?

Answer 34

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

Answer 35

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

Answer 36

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 :)

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

So y is not undefined?

Answer 43

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

Answer 44

Great. :)

Answer 45

One more thing, how do I show which is the minimum point and which is maximum? Is it by the second derivative?