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How to show that f(x) = x^3 - x^2 + x - 1 is never decreasing

  1. Jul 31, 2008 #1

    Air

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    1. The problem statement, all variables and given/known data
    Show [itex]f(x) = x^3-x^2+x-1[/itex] is never decreasing.


    2. The attempt at a solution
    [itex]f(x) = x^3-x^2+x-1[/itex]
    [itex]f'(x) = 3x^2 - 2x + 1[/itex]
    [itex]f''(x) = 6x - 2[/itex]


    3. The problem that I'm facing
    I don't understand what it means by never decreasing. Do I say that when differentiates, it gives a positive quadratic equation hence it never decreasing or when differentiated twice, it forms a positive linear equation hence never decreasing? :confused:
     
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  3. Jul 31, 2008 #2
    Re: Maxima/Minima

    For a function to be increasing, for any
    [tex]x_1<x_2[/tex];
    [tex]f(x_1)<f(x_2)[/tex].
    The opposite applies for decreasing functions.
     
  4. Jul 31, 2008 #3

    Air

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    Re: Maxima/Minima

    But, I'm only provided with one function so how can I show:

    [itex]f(x_1)<f(x_2)[/itex]

    What is [itex]x_1, x_2[/itex]?
     
  5. Jul 31, 2008 #4
    Re: Maxima/Minima

    [tex]x_1[/tex] and [tex]x_2[/tex] are simply two consecutive numbers.

    It's just logic, take [tex]x_1=2[/tex] and [tex]x_2=3[/tex];
    [tex]f(x_1)=5[/tex] and [tex]f(x_2)=20[/tex].

    ... and you can prove this for any two numbers, aslong as they are consecutive.
     
  6. Jul 31, 2008 #5

    Air

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    Re: Maxima/Minima

    So, I substitute two consecutive numbers into the function and show that it's increasing.

    No differentiation is required for this question?
     
  7. Jul 31, 2008 #6
    Re: Maxima/Minima

    Indeed. Think about it, if for any two consecutive numbers, their corresponding function value is greater, how can the function be decreasing?

    You could also use the theorem that on any interval (a,b), if f '(x) is > 0, then the function is increasing on that range.
    Take the interval (1,2) for example; f '(1.5) > 0 and this can be proven for all intervals for this function.

    This method is more useful when you are asked questions such as "Find the values of x where the function is increasing or decreasing", as you can use the derivative to find the critical points of the function and evaluate the sign of the derivate over each of the intervals.

    Hope this makes sense.
     
  8. Jul 31, 2008 #7

    HallsofIvy

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    Re: Maxima/Minima

    You have
    [tex]f'(x) = 3x^2 - 2x + 1= 3(x^2- (2/3)x+ 1/9)- 1/3+ 1= (x-1/3)^2+ 2/3[/tex]
    which is never 0. Checking x= 0, f'(0)= 1> 0 so f' is positive for all x.
     
  9. Jul 31, 2008 #8

    Defennder

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    Re: Maxima/Minima

    If the function never decreases, then it's derivative must be always be positive right? How can you show that is true for f'(x)?
     
  10. Jul 31, 2008 #9

    Dick

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    Re: Maxima/Minima

    YES. USE DIFFERENTIATION. You've already said the correct thing. f'(x) is a positive quadratic. If the derivative is always positive, a function is increasing. 'never decreasing' would just mean 'always increasing or constant'.
     
  11. Jul 31, 2008 #10

    Air

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    Re: Maxima/Minima

    Thanks. I understand! :smile:
     
  12. Jul 31, 2008 #11

    HallsofIvy

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    Re: Maxima/Minima

    Strictly speaking, if a function is never decreasing, it's derivative is never negative. That's not the same as saying "always positive" (although in this problem the derivative is always positive).

    for example, f(x)= x3 is never decreasing but f'(x)= 3x2 which is NOT always positive: f'(0)= 0.
     
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