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Homework Help: Finding critical numbers

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Find all critical numbers of:
    g(x)= sqrt(x2-4)
    f(x)= (1)/(x2-9)

    2. Relevant equations

    3. The attempt at a solution
    1) sqrt(x2-4)
    simplified to x(x-2)(x+2)-1/2 (please check)
    and got zeroes as x=0, x=2, x=-2
    and I got confused because if you do g(2) and g(-2) it equals zero. For some reason I can't tell if they are defined or undefined. x=0 works, so that is a critical number. The other two are throwing me off.

    2) (1)/(x2-9)
    zeroes were x=-3, x=3, x=0. Plugging 3 and -3 back into f(x) gave me undefined, so I'm pretty sure 0 is the only critical number.

    If you could please check the second and help with the first that would be great. Thank you
  2. jcsd
  3. Mar 22, 2010 #2


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    "1) sqrt(x2-4)
    simplified to x(x-2)(x+2)-1/2"

    - is that supposed to be the derivative of [tex] \sqrt{\,x^2 -4}[/tex]? if so, it isn't correct.

    what is your definition of a critical value? (writing it out can help you see the appropriate path)
  4. Mar 22, 2010 #3
    I'm honestly not sure what I did there... I re-did my derivative and found


    which would go to x/(((x+2)(x-2))1/2)

    and give zeroes of -2,2,0. -2 and 2 would make the denominator 0 and be undefined, leaving 0 as the only CP. Sound right? Sorry if any errors I did that really quick.
  5. Mar 22, 2010 #4


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    yes, the derivative is

    \frac{x}{\sqrt{\, x^2 - 4}}

    and this is zero or undefined for [itex] 0 \text{ and } \pm 2 [/itex].

    Again, what is your definition of a critical number? (same as critical value)
  6. Mar 22, 2010 #5
    Definition is:
    x=c is a critical number for f(x) if f(c) is defined and f'(c)=0 or f'(c) is undefined.

    So that means 2 and -2 WOULD be CPs?
  7. Mar 22, 2010 #6


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  8. Mar 22, 2010 #7
    And 0 would not be one because it is undefined in f(x) (sqrt of a negative makes it undefined), correct?

    And for the second problem, zero was the only one defined in f(x). Therefore it IS indeed a CP.
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