Expert: Paul Klarreich - 5/24/2007

Question
Find the critical numbers of the function

G(x) = X^1/3  - X ^-2/3

G(x) = ¡Ìt (1-t)

Find the absolute maximum and absolute minimum values of F on the given interval

F(x) = (X^2 ¨C 1)^3,   [-1, 2]


Answer
Questioner:   ted
Question:  Find the critical numbers of the function

G(x) = X^1/3  - X ^-2/3

G(x) = ¡Ìt (1-t)

Find the absolute maximum and absolute minimum values of F on the given interval

F(x) = (X^2 ¨C 1)^3,   [-1, 2]
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Hi, Ted,

Critical numbers means:
A. Values of x where  f'(x) = 0.  [Stationary point.]
B. Values of x where  f'(x) is undefined. [Singular point.]
C. Endpoints of an interval.


For:  G(x) = X^1/3  - X ^-2/3

You don't have any interval specified, so it's only A and B.

G'(x) = (1/3)x^-2/3  - (-2/3)x^-5/3

Simplify that a bit:
            1         2
G'(x) =  -------- + --------
        3 x^2/3    3x^5/3

Do a little algebra:

            1         2
G'(x) =  -------- + --------
        3 x^2/3    3x x^2/3

         x  +  2
G'(x) =  ---------
         3x x^2/3

Now  G'(0) = 0, so that is a critical (singular)  point.
Set the numerator = 0:  x + 2 = 0  gives  x = -2. [Stationary point.]
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G(x) = ¡Ìt (1-t)

Sorry, this did not come through.  Do not use special characters on this site -- they just get lost.  Find some other way.

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F(x) = (X^2 ¨C 1)^3,   [-1, 2]

Same comment.  Of course, here,  x = -1 and x = 2  will be critical points because they are end points.