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Show that a function is increasing

  • Thread starter Parthalan
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  • #1
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Hi,

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.

Homework Statement


Show that [itex]f(x) = 9x^2 + 3x[/itex] is strictly increasing on the interval [itex](0, 10][/itex]

Homework Equations




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:

Answers and Replies

  • #2
arildno
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What sign does f(v)-f(u) have?
 
  • #3
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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?
 
  • #4
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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.
 
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  • #5
Dick
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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.
 
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  • #6
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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
 
  • #7
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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].
QED.

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

Many thanks to the people who have helped.
 
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  • #8
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: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.
 
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  • #9
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Thanks!

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).
 
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  • #10
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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?
Don't forget, this is the pre-calculus section! :tongue:
 

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