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Edited by SpeechlessMoon: 11/6/2013 4:49:35 PM
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Math buffs, I have a calculus question

So in calculus we have been doing absolute extrema and critical points and all that jazz, and it has raised a question. I had a question on my homework that asked what the maximum and minimum of a line would be on the closed interval [0,2], where f(x) = 7/4X - 3. The minimum was -3 and the max was 1/2. I then asked what the minimum was on the interval (0,2]. I said it was negative infinity, but it was counted wrong. According to my teacher, the minimum did not exist. My reasoning for answering negative infinity is because technically the answer was - 2.9999... because the value would just keep approaching -3. Since infinity holds all numbers, I assumed that the value was infinite, because you could just keep adding 9's to the end of the decimal, and since the number was decreasing in value, I said negative infinity. If someone could shed some light on why I am wrong, that would be great, because I am currently in an argument with my teacher about it. [spoiler]Ahhh... it is nice to post on my computer again, it has been a while[/spoiler]

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  • Edited by Bolt: 11/6/2013 6:13:37 PM
    I had to read that three times before I caught it. For anyone who's still scratching their head: [0,2] (0,2] [0,2] [b]([/b]0,2] [0,2] [b][u]([/b][/u]0,2] It's a weird question, but negative infinity is definitely not the answer. In order to give you a visual sense of the reason why, look at the image I linked. It shows an example of a function where you could consider one of the minimums to be negative infinity (as you approach x=2 from the left). [url=http://www.wolframalpha.com/input/?i=%287%2F4%29x-3]Here[/url] is what your function looks like. Clearly your function doesn't approach negative infinity at 0. The error with your reasoning is that, while yes you can add an infinite number of nines after the decimal, you need to realize that each nine you add is an order of magnitude less than the one before it. You'll never add up (...down?) to negative three, much less negative infinity. As a general rule of thumb, infinity won't show up as a limit or value unless the function has the potential to divide a value by zero. With your function, since the only x is in the numerator, there is no real numerical (infinity isn't a number, it's a cardinal set) value of x for which the function can be undefined. With the function in the image I linked, when x equals 2, the denominator becomes zero, and you wind up with [b](1/0)[/b]+4, which is undefined. In general, if x isn't in the denominator, you really shouldn't have to worry about infinity. Hope that helped.

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    • -2.99999... Does not equal negative infinity, it equals -3, that is literally the first thing you learn in calc. Limits

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    • Edited by NaLo: 11/6/2013 6:04:05 PM
      It's -2,999... which is not negative infinity. It's just infinitely close to negative three. However one could argue that -2.999... = -3 [quote][i]lim x→0+ [/i] f(x)=(7/4)x - 3 = -3[/quote] But the above statement only says that it approaches 3. It will never become -3 though. And infinity doesn't necessarily hold all numbers. Here is an infinite non-repeating sequence using only two numbers for example. 101100111000...

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      • no u

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      • potato

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      • Hey, Calculus Tutor here. I just wanted to clarify exactly what the equation in the problem was. Could you add some parenthesis to clearly indicate what's in the denominator of the fraction? Your reasoning definitely seems off, but I'd like to get the big picture for what the problem was before I try to explain it.

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        • Edited by Jotaro Kujo PhD: 11/6/2013 5:49:13 PM
          I'm pretty sure that -2.999... isn't considered undefined like negative infinity. It has a "known" value between -2.999 and -3. I'm not taking BC until next year, so I'm just guessing.

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        • ... What??

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          • Bass.

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