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Find MIN

  1. Nov 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Explain why f(x) = 1/x must have a minimum on [1,2] but not on [-1,2].


    2. Relevant equations



    3. The attempt at a solution

    I don't know what to do. First, I took the x values from the given points and plugged them back into the function but the results do not coincide with the y values. I don't really know what to do.

    I also thought about differentiating it and then f'(x) = 0.
    I used the quotient rule:
    (x)(1)' - (1)(x)'/(x)^2
    -1/x^2 -> f'(x)=0
    -1/x^2= 0
     
  2. jcsd
  3. Nov 11, 2012 #2

    vela

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    [1,2] is the interval from 1 to 2, not the point (1,2).
     
  4. Nov 11, 2012 #3
    I still don't know how to find the answer.
    I've had exercises where I was able to write my answer as "MIN: (-1,-4) and (2,-4)" or "MIN: (0,0) and (3,0)" for instance and they were correct.
     
    Last edited: Nov 11, 2012
  5. Nov 11, 2012 #4

    Mark44

    Staff: Mentor

    So what does this tell you?

    One of the first things you should have done is sketch a graph of the function f(x) = 1/x. You would be able to see from the graph why there is a minimum on the interval [1, 2] and why there isn't one on the interval [-1, 2].
     
  6. Nov 11, 2012 #5
    what I don't understand is how would the graph of [1,2] and [-1,2] differ from [-1,-4] and [2,-4], for instance, as I found in one of my previous homework.
     
  7. Nov 11, 2012 #6

    Mark44

    Staff: Mentor

    This makes no sense. You aren't graphiing [1, 2] etc. These are intervals.

    You didn't answer my previous question. You found that if f(x) = 1/x, then f'(x) = -1/x2. Then you set f'(x) to 0, getting
    -1/x2 = 0

    What is the solution to this equation?
    What does that tell you about the graph of f(x)?
    Have you graphed f(x) = 1/x?
     
    Last edited: Nov 11, 2012
  8. Nov 11, 2012 #7
    there's no solution.
     
  9. Nov 11, 2012 #8

    Mark44

    Staff: Mentor

    Right. So what about the other questions I asked...
     
  10. Nov 12, 2012 #9

    HallsofIvy

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    The crucial point is that f(0) is not defined. That matters on [-1, 2] but not on [1, 2]. Do you see why?

    There is a theorem that says if a function is continuous on a closed and bounded interval it has both maximum and minimum values on the interval. Do you see why that applies to [1, 2] but not to [-1, 2]?

    Finally there is a theorem that says that if a function is continuous on a closed and bounded interval and differentiable in its interior, then the max and min occur where the derivative is 0 or at the end points.
     
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