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Show that a function is increasing

  1. Mar 16, 2007 #1

    I'm in need of some help putting together this proof. I'm not sure if it's just harder than it looks, or I need to spend more time learning how to put them together.

    1. The problem statement, all variables and given/known data
    Show that [itex]f(x) = 9x^2 + 3x[/itex] is strictly increasing on the interval [itex](0, 10][/itex]

    2. Relevant equations

    3. The attempt at a solution
    I realise there are various arguments involving the turning point of a quadratic equation, but I don't think this is the type of answer that they're looking for.

    Let [itex]u < v[/itex] be two elements of the interval [itex](0, 10][/itex]. Then from [itex]f(u) < f(v)[/itex] we obtain:
    [tex]9u^2 + 3u < 9v^2 + 3v[/tex],
    [tex]9u^2 + 3u - 9v^2 + 3v < 0[/tex]

    From here, I think I need to get to [itex]u < v[/itex], but I can't see how to do it.

    Thanks for any help.
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 16, 2007 #2


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    What sign does f(v)-f(u) have?
  4. Mar 16, 2007 #3
    One thing you can do is evaluate the first derivative of the function on that specified interval.
    Do you know how can you interpret the sign of the derivative of a function over a specified interval on how the function itself is behaving on that interval?
  5. Mar 16, 2007 #4
    The sign must be [itex]f(v) - f(u) > 0[/itex], otherwise [itex]f(v) < f(u)[/itex].

    Also, thanks for the suggestion about using the derivative, but I'm wondering if this can be done without.

    Now for another go.

    Let [itex]u < v[/itex] be in the interval [itex](0, 10][/itex]. Suppose that [itex]f(v) > f(u)[/itex]. Then:
    [tex]f(v) - f(u) < 0[/tex]
    [tex]9v^2 + 3v - 9u^2 + 3u < 0[/tex]
    Which can only be true for [itex]u > v[/itex], hence, a contradiction.

    Is that it, or did I miss a step?

    Thanks for your help.
    Last edited: Mar 16, 2007
  6. Mar 16, 2007 #5


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    Your reasoning is circular. If you really want to show its increasing without derivatives show f(v)-f(u) is positive by factoring the expression for it. Hint: it's divisible by (v-u). Correct the sign on the 3u term first.
    Last edited: Mar 16, 2007
  7. Mar 16, 2007 #6
    Parthalan - read carfully what Dick is suggesting.

    As an added hint start of your reasoning like this

    Let u, v be two elelements in the interval (0,10] with u<v.

    f(v)-f(u) = ................

    and then as Dick says use the information you have to show this must be >0
  8. Mar 17, 2007 #7
    I'm sure I have you all banging your head against a wall by now.

    I got this far:

    Let [itex]f(x) = 9x^2 + 3x[/itex] and [itex]u, v \in (0, 10][/itex], where [itex]u < v[/itex]. If the function is strictly increasing, then [itex]f(u) < f(v)[/itex]. From this, we determine:
    [tex]f(v) - f(u) = (9v^2 + 3v) - (9u^2 + 3u)[/tex]
    [tex]= 9v^2 + 3v - 9u^2 - 3u[/tex]
    [tex]= (-3u + 3v)(1 + 3u + 3v)[/tex]
    [tex]= -3(u - v)(1 + 3u + 3v)[/tex]
    Since [itex]u < v[/itex], then [itex]u - v[/itex] must be negative. Since the product of two negatives is always a positive, and [itex](1 + 3u + 3v)[/itex] will also be positive, then [itex]f(v) - f(u)[/itex] must be positive, and therefore, [itex]f(u) < f(v)[/itex].

    Is that right, or am I still going in circles?

    Many thanks to the people who have helped.
    Last edited: Mar 17, 2007
  9. Mar 17, 2007 #8
    :uhh: Is it even necessary to write so much about this?

    If [tex]0< x_{1} < x_{2}[/tex]

    [tex]3x_{1} < 3x_{2}[/itex] and [tex]9{x_{1}}^2 < 9{x_{2}}^2 [/tex] . Therefore

    [tex]9{x_{1}}^2 + 3x_{1} < 9{x_{2}}^2 + 3x_{2}[/tex]. We're done here.
    Last edited: Mar 17, 2007
  10. Mar 17, 2007 #9

    That was what I tried to do in the first place, but I forgot to factor it first, and ended up trying to show that it was increasing because the difference between [itex]f(x_2)[/itex] and [itex]f(x_1)[/itex] was positive. Changing the inequality directly seems to be the easier way (not as easy as using the derivative, but unfortunately, it's a course that comes before calculus).
    Last edited: Mar 17, 2007
  11. Mar 17, 2007 #10
    Don't forget, this is the pre-calculus section! :tongue:
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