Expert: Paul Klarreich - 12/3/2008

Question
Identify the critical point and determine the local extreme values...
y=x(4-x^2)^(1/2)
I tried working this out on my own and with my dad.  First, I tried to find the derivative and set it = to zero.  I used the quotient rule, and got u=x, u'=1, v=(4-x^2)^(1/2), v'=((4-x)^(-1/2))/2.  I got: y'=(4-x^2)^(1/2)-x/2(4-x^2)^)^(1/2) all divided by (4-x^2).  When I simplified it all out, I got y'=(2*(4-x^2)-x)/(2(4-x^2)^(1/2)*(4-x^2)).  But then I got confused as to how to set that = to zero and find critical points.  I think you can multiply out the top and get a quadratic, which would be y'=-2x^2-x+8.  When I use the quadratic formula, I get x=-2.26557 and 1.76556.  When I plug those numbers back into the original equation, I get (-2.26556, -6.8297) and (1.76556, 1.65887).  However, my answer book says I should be finding x to equal 2, -2, -(2)^(1/2), and 2^(1/2).  I'm not sure how they got those answers, so I'd really appreciate it if you helped me out.  Thank you!

Answer
Questioner:   Grace
Category:  Calculus
Private:  No
 
Subject:  Max-min critical points
Question:  Identify the critical point and determine the local extreme values...
y=x(4-x^2)^(1/2)


I tried working this out on my own and with my dad.  First, I tried to find the

derivative and set it = to zero.  


I used the quotient rule,

>> Why did you do that?  It looks like a product to me.  Did you mean to write:

         x
y = -------------
   (4 - x^2)^1/2


I think you just misread it.  If the example is as you wrote it -- a product, then the book's answers are correct.

.......................

Let me know right away.  And PLEASE, don't write as if this is a

stream-of-consciousness.  The English language has things called PARAGRAPHS.

..............................................


and got u=x, u'=1, v=(4-x^2)^(1/2), v'=((4-x)^(-1/2))/2.  I got:

y'=(4-x^2)^(1/2)-x/2(4-x^2)^)^(1/2) all divided by (4-x^2).  When I simplified it all

out, I got y'=(2*(4-x^2)-x)/(2(4-x^2)^(1/2)*(4-x^2)).  


But then I got confused as to how to set that = to zero and find critical points.  I

think you can multiply out the top and get a quadratic, which would be y'=-2x^2-x+8.  

When I use the quadratic formula, I get x=-2.26557 and 1.76556.  When I plug those

numbers back into the original equation, I get (-2.26556, -6.8297) and (1.76556,

1.65887).  However, my answer book says I should be finding x to equal 2, -2,

-(2)^(1/2), and 2^(1/2).  I'm not sure how they got those answers, so I'd really

appreciate it if you helped me out.  Thank you!


y' = (x)(-x/sqrt(4 - x^2)) + (1)(sqrt(4 - x^2))

      - x^2
y' = --------- + sqrt()
      sqrt()

    - x^2 + 4 - x^2
y' = -------------------
           sqrt()


    - 2x^2 + 4
y' = ------------
        sqrt()


x = +- sqrt(2), +- 2

(singular pts are critical points, too.)