inquiry based on the noise of graphing software

johnnytheflipper · 2008-01-13 05:25:19

I recently graphed a complicated function on geometer's sketchpad, in hopes that i could determine a local minimum graphically, seeing as that is what the question was asking. I used the trace feature to narrow in on the point I wanted to find, and as I narrows in on that point, I found that there were 2 x-values that were infinitely close to eachother (for example, .38798, and .39124) that produced the same minimum y-value 0.15. I am sure that there is not just a straigh line between these two points seeing as I am dealing with a quartic function.

My question is, Do i take the average between these two points (i dont think so, because it still wouldnt be acturate), or do I state that "the minimum y-value is located Somewhere between these two x-values". I dont want to make this statement because the question specifies to determine a specific x-value graphically.

Thank you very much,

Johnnytheflipper

LuisRodg · 2008-01-13 05:57:09

Would you care to post such function that you graphed?

Maybe using calculus you can find the minimum value without having to graph it. Minimum and maximum values occur only at the critical points of the graph where the derivative equals 0 or the derivative doesnt exist or they can also occur at the endpoints of a continuous function in a closed interval.

Last edited by LuisRodg (2008-01-13 05:57:54)

johnnytheflipper · 2008-01-13 06:01:00

I am not taking calculus so I have no idea what you just said haha. I have that course next semester. My question was clear. I have to use graphing software to show this. But there is noise. What do i do?

LuisRodg · 2008-01-13 06:07:09

Im not sure what to tell you. As I said before, can you post the function so I can graph it myself and see?

johnnytheflipper · 2008-01-13 06:31:52

okayy f(x) = (((40+3x+x^2)/500)-0.1)^2 + 0.15 I must find the minimum value between x=0 and x=4. Do not let the straight line appearance fool you. IT IS NOT a straight line.

LuisRodg · 2008-01-13 06:44:11

Ok your right. it DOES look like a straight line. I kept changing the windows settings on my calculator but it still looked as a line. However, my TI-89 calculator has a function that finds the minimum value of a function on an interval, I did it from x=0 to x=4 and it says the minimum value is at x=2 and y=0.15

johnnytheflipper · 2008-01-13 07:08:44

I am achieving the same result exept x= 1.9987 and x=2.0077.... what should i do? if i round to 2, will he deduct me?

LuisRodg · 2008-01-13 07:14:20

The real answer is 2. You are getting 1.9987 and 2.0077 because of inaccuracy with the graphing. I would answer 2.

luca-deltodesco · 2008-01-13 07:16:05

quote it to how many significant figures it is consistent.

i.e. for you its 3 figures. 1.9987 to 3sf is 2.00 and 2.0077 to 3sf is also 2.00 so you can safely quote that the root is at x = 2.00 to 3 significant figures

johnnytheflipper · 2008-01-13 07:16:42

thank you. surprizingly, I found help in a forum for once. I will continue using this forum for my mathematical inquiries. Again, thank you very much LuisRodg.

mathsyperson · 2008-01-13 08:16:28

I'm probably jumping in too late here, but I just wanted to say that you can evaluate the minimum of this one analytically, rather than looking at its graph. I'd say it's always better to do that if you can.

The function is in the form of [something]^2 + 0.15.
[something]^2 will always be non-negative, so it will be minimal when [something] = 0.

So now we try to solve (40+3x+x^2)/500)-0.1 = 0.
If an x exists that solves this then we're done, if not then things get a bit more complicated.

Rearranging:

(40+3x+x^2)/500 = 0.1
40+3x+x^2 = 50
x^2+3x-10 = 0

Factorise: (x+5)(x-2) = 0

Hence, x = 2 or -5.

Ignore -5 since it's not in the range we're looking at, and so we conclude that the function is minimal when x=2.

At that point, the big thing that all gets squared is 0, and so the function is simply equal to 0.15.

Math Is Fun Forum

#1 2008-01-13 05:25:19

inquiry based on the noise of graphing software

#2 2008-01-13 05:57:09

Re: inquiry based on the noise of graphing software

#3 2008-01-13 06:01:00

Re: inquiry based on the noise of graphing software

#4 2008-01-13 06:07:09

Re: inquiry based on the noise of graphing software

#5 2008-01-13 06:31:52

Re: inquiry based on the noise of graphing software

#6 2008-01-13 06:44:11

Re: inquiry based on the noise of graphing software

#7 2008-01-13 07:08:44

Re: inquiry based on the noise of graphing software

#8 2008-01-13 07:14:20

Re: inquiry based on the noise of graphing software

#9 2008-01-13 07:16:05

Re: inquiry based on the noise of graphing software

#10 2008-01-13 07:16:42

Re: inquiry based on the noise of graphing software

#11 2008-01-13 08:16:28

Re: inquiry based on the noise of graphing software

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