
#1
Mar1607, 07:54 AM

P: 27

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. 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. 



#2
Mar1607, 07:58 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

What sign does f(v)f(u) have?




#3
Mar1607, 08:01 AM

P: 239

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
Mar1607, 08:17 AM

P: 27

Show that a function is increasing
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. 



#5
Mar1607, 09:01 AM

Sci Advisor
HW Helper
Thanks
P: 25,171

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 (vu). Correct the sign on the 3u term first.




#6
Mar1607, 11:38 AM

P: 123

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
Mar1707, 03:45 AM

P: 27

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. 



#8
Mar1707, 07:29 AM

P: 1,520

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. 



#9
Mar1707, 07:44 AM

P: 27

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). 



#10
Mar1707, 02:45 PM

P: 291




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