darkrider2001 Posted January 12, 2007 Report Share Posted January 12, 2007 he question was: Rolle's Theorem:f(x) = x3 + x2 [-1, 1](It had some other stuff about finding all x values where the theorem applies) My work: f`(x) = 3x2 + 2x f(x) is continuous on interval [-1,1]andf`(x) is continuous on interval (-1,1) I know f(-1) != f(1) (not equal) but I did this, f(x)=x3 + x2 = x2(x + 1) = 0x=0; x=-1 0 and -1 are inside the continuous interval [-1,1]andf(0) = f(-1) = 0 So Rolle's theorem applies! Now we know there is some point on [-1,1] where f`(x)=0 f`(x) = 3x2 + 2x = x(3x + 2)x=0 ; x=-2/3 Ok that was pretty much the work I had on my paper. My friend said it does not apply because f(-1)!=f(1) (not equal) and he told me he got the problem right (he did the quiz the other week) I just did this quiz this afternoon after school, so I haven't gotten it back yet. I'm sure it's right as I swear I saw one of the sources I'm studying from do a similar problem in almost the exact same way. However, tomorrow is the last day of class, and I'm unsure if she'll check it right. I think it's right, but in case if she checks it wrong, can someone give me some kind of proof as to why it's right. Much thanks to anyone who helps as this is very important for my grade. Quote Link to comment Share on other sites More sharing options...

darkrider2001 Posted January 12, 2007 Author Report Share Posted January 12, 2007 sorry i think i posted this in the wrong place. Quote Link to comment Share on other sites More sharing options...

darkrider2001 Posted January 12, 2007 Author Report Share Posted January 12, 2007 can anyone help me i have to go to bed soon and this is due tommorow. Quote Link to comment Share on other sites More sharing options...

CraigD Posted January 12, 2007 Report Share Posted January 12, 2007 …My friend said it does not apply because f(-1)!=f(1) (not equal) and he told me he got the problem right (he did the quiz the other week)…I think it's right, but in case if she checks it wrong, can someone give me some kind of proof as to why it's right.…Your friend is right – Rolle's theorem doesn’t apply to a function continuous on [math]\left[-1,1\right][/math] and differentiable on [math]\left(-1,1\right)[/math] unless [math]f(-1)=f(1)[/math]. The point of Rolle’s theorem is that it isn’t necessary to actually know the formula of [math]f()[/math], only that it’s continuous and differentiable on the interval and that its end values are equal. Even though the theorem applies to intervals covered by [math]\left(-1,1\right)[/math], it doesn’t apply to the given interval. Put in another way, if, rather than being given [math]f(x) = x^3 +x^2[/math], you were only given [math]f(-1)=0[/math], [math]f(1)=2[/math], and [math]f()[/math] is continuous on [math]\left[-1,1\right][/math] and differentiable on [math]\left(-1,1\right)[/math], you wouldn’t be able to use Rolle’s theorem to conclude [math]f’(x)=0[/math] for [math]-1<x<1[/math]. Giving you more information than is applicable to the theorem was, I fear, a trick. The only suggestion I can make is that you convince your teacher that you understand the theorem, but misunderstood the question, and went beyond what it was asking in your answer. Good luck. Quote Link to comment Share on other sites More sharing options...

darkrider2001 Posted January 12, 2007 Author Report Share Posted January 12, 2007 wow thanks i get it now Quote Link to comment Share on other sites More sharing options...

Qfwfq Posted January 12, 2007 Report Share Posted January 12, 2007 Well, I can only suppose what the excercise was asking but what you did is, at least, on the right track. You found an interval inside the one given, which fits the theorem's hypothesis, but you didn't quite show that the thesis holds in that subinterval (unless you pointed out that -2/3 is between -1 and 0). I can only guess what the teacher was after, according to how difficult the test is meant to be. There are certainly many other good intervals besides [-1, 0] so a more sophisticated solution would be to describe the full set of them. If the quiz asked for "at least one" of them then you'd be OK if you made it clear enough that you understood the above point about -2/3 being inside the interval you found. Quote Link to comment Share on other sites More sharing options...

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