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Find the extreme values

  1. Dec 12, 2004 #1
    Hi everyone!

    Find the extreme values of the function and where they occur. I have to do this without graphing them and without a calulator.

    1.) y= sqrt( x^2 - 1 )
    y'= [.5( x^2 -1)^ (-3/2)][2x]
    y'= x / [(x^2 - 1)^(3/2)]
    0 = x / [(x^2 - 1)^(3/2)]
    Now what???? I can't solve for x?

    2.) y= (x^2 - 1)^-1
    y' = -1 (x^2 -1) ^-2
    0 = -1(x^2 - 1)^-2
    x= -1 and 1?????
    Is this correct? If so how do I know which is the min and which is the max?

    3.) y= 1/ sqrt(1- x^2)
    y'=-1(sqrt(1 -x^2))^-2
    0 = -1/ (1- x^2)
    x = 1 and -1??
    Is this correct? If so how do I know which is the min and which is the max?

    ~Thanks
     
  2. jcsd
  3. Dec 12, 2004 #2

    quasar987

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    Finding the extrema is not the same as finding the minuma an maxima. An extremum of the function is simply a point OF THE DOMAIN for which the derivative is zero. So step one is to find the domain of the functions, keeping in mind that negative roots and division by zero are not defined. This means that for a particular function, all real numbers x for which either of these two cases arise are excluded from the domain of the function.

    You also want to check on your derivation rules some more as there are mistake in all 3 problems.

    Reminder: for all x, 1/x = x^-1

    Hint: the answer for all the first two problems is that there are no extrema and there is one in the third.
     
    Last edited: Dec 12, 2004
  4. Dec 12, 2004 #3

    dextercioby

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    The last 2 differentiations were done wrong.And the first one,which was correct,should have lead to the obvious result "x=0".
    So,make the calculations again,if u know how. :wink:
     
  5. Dec 12, 2004 #4

    quasar987

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    As far as I can see, the first differentiation is not correct either.

    [tex]y= \sqrt{x^2-1} \Rightarrow y' = \frac{1}{2\sqrt{x^2-1}} (2x) = \frac{x}{\sqrt{x^2-1}} [/tex]

    because [itex]1/2 - 1 = -1/2 \neq -3/2[/itex]


    The result is also x = 0, as is obtained by multiplying both sides by [itex]\sqrt{x^2-1}[/itex] after setting y' = 0, but x = 0 is to be rejected as it is not a value of the domain of the function since

    [tex]f(0) = \sqrt{0^2-1} = \sqrt{-1} [/itex]

    and like I mentionned, negative roots are not defined in [itex]\mathbb{R}[/itex].
     
    Last edited: Dec 12, 2004
  6. Dec 12, 2004 #5

    dextercioby

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    Yes,u're right,i didn't look too carefully and didn't see that "3/2". :tongue2: As for the domains for the functions themselves,the fact he had no idea how to differentiate seemed a greater problem to me.But you're right here,as well,since the first step when plotting a function is to determine its domain.Anyhow,he lacks knowledge,and my power to concentrate is not that big. :yuck: I should have seen that "3/2",damn it!!!

    Daniel.
     
  7. Dec 12, 2004 #6
    What's the difference in finding the min and max, and the extreme values?

    2.) y= (x^2 - 1)^-1
    y= (x^2 - 1)^-1
    y'= [-1(x^2 -1)^-2] (2x)
    y'= (-2x)/[(x^2 -1)^2]
    x = 0

    y= (x^2 - 1)^-1
    y= (0-1)^-1
    y= -1

    Why is there no extreme value??? Did I do something wrong?

    3.) I got x= 0 so would the answer be (0,1)?
     
    Last edited: Dec 12, 2004
  8. Dec 12, 2004 #7

    quasar987

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    No that's good, I made the mistake, sorry.

    The extreme values are CANDIDATES for being a maximum or a minimum of the function. They are simply those points of the domain for which the derivative is 0. After finding all the extreme values, we may then proceed to determining wheter each is a minimum of the function, a maximum, or neither. If it is the later case, we call the point a saddle point (because the saddle point of a function of two variables looks like a saddle when we plot the graph :smile:)

    err... I couldn't draw the thing nor find a picture of a saddle point of a function of one variable on the internet but for an exemple, plot the graph of [itex]f(x)=x^3[/itex]. x = 0 is a saddle point.
     
    Last edited: Dec 12, 2004
  9. Dec 12, 2004 #8
    1.) the answer is there is no extreme value?

    2.) the answer is the extreme value is at x=0

    3.) the answer is the extreme value is at x=0?

    So the extreme value is the x value of the max or min? How do you find out if it is a max or min?
     
  10. Dec 12, 2004 #9

    quasar987

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    Yes, but it is convential to simply denote the extreme values of a function as their value in x. So the answer would not be (0,1) but simply 0.

    Right.



    You find out if it is a max or min by examining the derivatives of higher order. The method should be explained in your textbook.
     
  11. Dec 12, 2004 #10
    My teacher said i can find the min and max with the 1st der and the points of inflection with the 2nd deriv. How would I do this? How can I find the min and max with the 1st deriv if the 1st deriv is used to find the extremes?
     
  12. Dec 12, 2004 #11

    quasar987

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    The genral method:

    1) list all your extrema and order them from smaller to bigger

    2) find the SIGN of f '(x) between each extrema.

    Conclude as follow: Let c be an extremum.

    i) If the sign of the derivative is positive to the left* of c and negative on the right, c is a maximum.

    ii) If the sign of the derivative is negative to the left of c and positive on the right, c is a minimum.

    iii) If the sign is the same to the left and to the right, c is a saddle point.


    * The expressions "to the right" and "to the left" should be understood as follow: if c is the smaller of all the extrema, then "to the left" means "in the interval [itex]]-\infty,c[[/itex]". If c is the larger of all the extrema, then "to the right" means "in the interval [itex]]c,-\infty[[/itex]". If c is in-between two extrema, then "to the left" means "in the interval [itex]]b,c[[/itex]" (where b is the extremum smaller than c) and finally, "to the right" means "in the interval [itex]]c,d[[/itex]" (where d is the extremum larger than c).


    Now if you don't get this, go ask your teacher.
     
  13. Dec 12, 2004 #12
    If it was that easy I would have done this first. She just gave us small notes on this stuff and that's it. She did not explain it or anything. She gave this to us for a homework assignment and we are going to have a quiz on it next class. Before we take the quiz we are not allowed to ask any questions. She will go over the problems AFTER the quiz.

    I think I understand now. Thanks for your help :smile: .
     
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