by maryk
Tags: theorem
 P: 4 Hi there, hopefully someone can help me I'm completely lost!! I'm trying to solve sin((pi*x)/6) >= (x/2) for o<= x <= 1 using the mean value theorem. I think I need to show that f ' (c) >= 0, however this f ' (c) is negative for some values in the interval? HELP!!
 P: 1,635 ok, if you are trying to prove that: $$sin\frac{x\pi}{6}\geq \frac{x}{2}, 0\leq x \leq 1$$ so let $$f(x)=sin\frac{x\pi}{6}- \frac{x}{2},$$ we notice that $$f(0)-f(1)=0$$ now we take any point on the interval $$0\leq x \leq 1$$ say $$x=\frac{1}{2}$$, then if f(x) is positive or negative, it actually will meant that f(x) is either positive or negative on the whole interval $$0\leq x \leq 1$$ $$f(\frac{1}{2})=sin\frac{\pi}{12}-\frac{1}{4}>0$$ so $$f(x)=sin\frac{x\pi}{6}- \frac{x}{2}>0=>sin\frac{x\pi}{6}\geq\frac{x}{2},$$ Well,i don't know whether this answers the requirements of the problem though. Sorry!
P: 588
 Quote by sutupidmath we notice that $$f(0)-f(1)=0$$ now we take any point on the interval $$0\leq x \leq 1$$ say $$x=\frac{1}{2}$$, then if f(x) is positive or negative, it actually will meant that f(x) is either positive or negative on the whole interval $$0\leq x \leq 1$$
This is not true in general, you also have to show that there are no zeros in [0,1]. It is not enough, that f vanishes at the end points.

P: 1,635

 Quote by Pere Callahan This is not true in general, you also have to show that there are no zeros in [0,1]. It is not enough, that f vanishes at the end points.
I think we can be sure that there are no zeros on [0,1] since the function f(x) does not change sign at the endpoints. Am i right?
From the IVT we know that if f(a)f(b)<0, than there exists a poing cE(a,b) such that f(c)=0. But since f(a)f(b) is not <0 in our case, can we safely assume that there is no point cE(a,b) such that f(c)=0 ?? I think we can.
 P: 588 I think it doesn't matter whether or not f "changes sign at the endpoints". You're right that f(a)f(b)<0 ensures the existence of an c in [a,b] with f(c)=0. The converse is not true. What you can probably easily prove is that f(a)f(b)<0 iff there is an odd number of zeros (counted with multiplicity) in [a,b]. A counterexample to your reasoning would be [a,b]=[0,10 pi], f(x)=sin(x), Then f(a)=0=f(b), but there is a bunch of zeros in [a,b].
P: 1,635
 Quote by Pere Callahan I think it doesn't matter whether or not f "changes sign at the endpoints". You're right that f(a)f(b)<0 ensures the existence of an c in [a,b] with f(c)=0. The converse is not true. What you can probably easily prove is that f(a)f(b)<0 iff there is an odd number of zeros (counted with multiplicity) in [a,b]. A counterexample to your reasoning would be [a,b]=[0,10 pi], f(x)=sin(x), Then f(a)=0=f(b), but there is a bunch of zeros in [a,b].
Oh,yeah, i see now!
 P: 4 Thanks for your help everyone, think I need to show that it is true for every point within 0<= x <= 1, not just the endpoints and a point in the middle of the interval. I was thinking the working should start as follows: f(x) = sin ((pi*x)/6) - (x/2) and we need to show this greater than or equal to zero. f ' (x)= (pi/6)cos((pi*x)/6) - 1/2 since f(0)=0, we can rearrange the mean value theorem in the form: f(x) = x f ' (c), where f ' (c) from above is (pi/6)cos((pi*c)/6) - 1/2 for some c between 0 and 1. Therefore if i can show f ' (c) is greater than or equal to 0 for c between 0 and 1, then f(x) will also be greater than or equal to zero since we will have x times this where x is between 0 and 1. However, we are working in radians and f ' (c) is negative for c=0.6, for example. Soooo confused any ideas?? I think this is a valid argument, dont know why it wont work!!!
 P: 1,635 ok here it is another try to justify that my claim in post #2 holds. Using the mean value theorem we get that:$$\exists c\in(0,1)$$ such that $$f'(c)=f(1)-f(0)=0$$ so now $$f'(x)=(sin\frac{c\pi}{6}- \frac{x}{2})'=\frac{\pi}{6}cos\frac{\pi x}{6}-\frac{1}{2}$$ so $$f'(c)=\frac{\pi}{6}cos\frac{\pi c}{6}-\frac{1}{2}=0=>cos\frac{\pi c}{6}=\frac{1}{2}$$ After we solve this we get for c $$c_1=\frac{6}{\pi}arccos\frac{3}{\pi}+2k\pi$$ and $$c_2=-\frac{6}{\pi}acrcos\frac{3}{\pi}+2k\pi$$ from here since $$c\in(0,1)$$ it means that there is only one critical point in the interval (0,1), so it also means that f'(x) changes sign only once in it, if c is actually a local min or max. Now by second derivative test we see that $$f''(c)<0$$ so it means that $$c=\frac{6}{\pi}arccos\frac{3}{\pi}$$ is a maximum. so now i think that my claim that f(x)>0 on the interval (0,1) holds.Because say that there would be another point k, on the interval (0,1) such that f(k)=0, then it automatically would mean that there must be another point say m, such that it should be a local min/max on that interva. But this actually contradicts the fact that there is only once critical point, and hence only one local max on the whole interval (0,1). Hence we have the desired result that: $$f(x)=sin\frac{x\pi}{6}- \frac{x}{2}>0=>sin\frac{x\pi}{6}\geq\frac{x}{2},$$ I think this approach should work.
 P: 4 though when solving for c do we not get c= (6/pi) arccos (1/2) ? which equals 2 and is therefore not contained in the interval?
 P: 4 Sorry, ignore that last message, I lost a coefficient! lol thank you v v much for your help

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