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Homework Help: Critical Number from Analying a Graph.

  1. May 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Find all critical numbers of f(x)= x[tex]\sqrt{2x+1}[/tex]

    2. Relevant equations

    Derivative, Product Rule

    3. The attempt at a solution

    x[tex]\sqrt{2x+1}[/tex] = 0
    x must be -1/2 & 0
    (-1/2, 0) Critical Point.
    (0, 0) Critical Point.

    First derivative:
    f'g + g'f
    (1)([tex]\sqrt{2x+1}[/tex])+(1/2[tex]\sqrt{2x+1}[/tex])(x)(2)
    =([tex]\sqrt{2x+1}[/tex]) + (x / [tex]\sqrt{2x+1}[/tex])
    LCD:
    (([tex]\sqrt{2x+1}[/tex]) ([tex]\sqrt{2x+1}[/tex]) + 1)) / ([tex]\sqrt{2x+1}[/tex])

    = 2x+1 / [tex]\sqrt{2x+1}[/tex]
    So y= 0 when x= -1/2 and 0
    (-1/2, 0) Critical Point.

    The second derivative I wouldn't post it her since it gets pretty messy. However, I found out that y=o for the second derivative must have x = -1/2

    Now the problem is that my answer for all critical points are wrong. What am I doing wrong? PPLEASE HELP!
     
  2. jcsd
  3. May 6, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    [itex]y=x \sqrt{2x+1}[/itex]

    [tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{2\sqrt{2x+1}}(2)[/tex]

    [tex]\frac{dy}{dx}= \sqrt{2x+1} + (x)\frac{1}{\sqrt{2x+1}}[/tex]

    [tex]\frac{dy}{dx}=\frac{(\sqrt{2x+1})^2 + x}{\sqrt{2x+1}}=0[/tex]
     
  4. May 6, 2008 #3
    OH! Thanks for that!
    But, the thing is.. the answer for the critical number is -1/3 and from looking at the first derivative, one of the critical point is -1/2.

    So what am I doing wrong?
     
  5. May 6, 2008 #4
    While the derivative is undefined at x = -1/2, is the original function differentiable there, and what does that tell you about whether -1/2 is a critical number? Otherwise, if x > -1/2, then you can solve for dy/dx = 0, and you should find x = -1/3. Try it again using the derivative rock.freak posted.
     
  6. May 7, 2008 #5
    OH! Thanks for that. Problem solved.
     
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