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 April 12th, 2012, 06:35 PM #1 Senior Member   Joined: Apr 2012 Posts: 112 Thanks: 0 Number theory and finding the domain and range! Hi I have three questions, two on number theory and the other on finding range and domain. 1.) Find the largest domain possible for the function f(x)-> (2)/(2x^2 -6x+5) and find the range corresponding to this domain. I would like to know, is there any easy, shortcut method to do this? I have tried, it took a long time and my answer came out incorrect. What I did was to manually sub in numbers for x and then derive a conclusion from there. It is tedious and I am aware that I cant do this in an exam. How would you solve it? 2.) a,b and c are consecutive terms of both an AP and GP. Therefore a=b=c I don't understand why a=b=c. Surely the value must increase or decrease from a to c? Unless they're talking about the difference, but I don't see that in the question. 3.) 1+((3/2)x)+((3/2)x)^2+((3/2)x)^3+... I know that the common ratio is r=(3/2)x The question asks, write down the range of values for x for the series to converge. The answer was -10, is that right? Thank you so much!
April 12th, 2012, 07:08 PM   #2
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Math Focus: Calculus/ODEs
Re: Number theory and finding the domain and range!

1.) Equate the denominator to zero to find what values (if any) to exclude from the set of reals for the domain:

$2x^2-6x+5=0$

$x=\frac{6\pm\sqrt{-4}}{4}$

Thus, we find no real numbers to exclude. Thus the domain is:

$$$-\infty,\infty$$$

Now to find the range, the function will have its maximal value when the denominator is at its minimum, which will occur at the axis of symmetry:

$x=-\frac{b}{2a}=-\frac{-6}{4}=\frac{3}{2}$

At this value of x the denominator is $\frac{1}{2}$ and thus $f_{\text{max}}=\frac{2}{\frac{1}{2}}=4$

Since the function has a horizontal asymptote of $y=0$, the range is then:

$\(0,4\]$

[attachment=0:44r3687v]range.jpg[/attachment:44r3687v]

2.) For a, b and c to be consecutive terms in an arithmetic progression, we require:

$b=a+d,\,c=a+2d$

For a, b and c to be consecutive terms in a geometric progression, we require:

$b=ar,\,c=ar^2$

Thus, we have:

$b=a+d=ar$

$c=a+2d=ar^2$

Hence:

$d=ar-a=\frac{ar^2-a}{2}$

$2(r-1)=(r-1)(r+1)$

$(r-1)(2-(r+1))=0$

$(r-1)^2=0$

$r=1\:\therefore\:d=0\:\therefore\:a=b=c$

http://en.wikipedia.org/wiki/Geometric_series
Attached Images
 range.jpg (9.7 KB, 121 views)

 April 12th, 2012, 08:41 PM #3 Senior Member   Joined: Apr 2012 Posts: 112 Thanks: 0 Re: Number theory and finding the domain and range! Thank you so much!!

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