Little confussed (Mean Value Theorem)

powp
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Hello All

I am a bit confussed with this question I have.

Show that the equation 2x - 1 - sin x = 0 has exactly one root. So this apears in the Mean Value Theorem section of my book. If some one can help it would be great.

I believe I need to use the Intermediate Value Theorem to show that a root exists, but am unsure of what values to use for it. Do I just pick random numbers? I need to show that there is a value between f(a) and f(b) that equals zero which will be a the root. Am I correct??

THanks
 
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The "intermediate value theorem" says that a continuous real function hits all the points in between two points in its range. To put it into human language, for example, it means that if right now you're 425 miles away from your baby, and two months ago you were 0 miles away from your baby, then, since 0 < 133 < 425, there must have been a time, in the last two months, that you were exactly 133 miles away from your baby.

Carl
 
powp said:
Hello All
I am a bit confussed with this question I have.
Show that the equation 2x - 1 - sin x = 0 has exactly one root. So this apears in the Mean Value Theorem section of my book. If some one can help it would be great.
I believe I need to use the Intermediate Value Theorem to show that a root exists, but am unsure of what values to use for it. Do I just pick random numbers? I need to show that there is a value between f(a) and f(b) that equals zero which will be a the root. Am I correct??
THanks

If x= 0 f(x)= 2x-1-sin x= -1<0. If x= \pi, f(x)= 2x- 1- sin x= 2\pi> 0. So there exist at least one root between 0 and \pi.
However, as carlB pointed out, that's the "intermediate value property", not the "mean value theorem".
Suppose there were more than one root, at, say, x1 and x2. Then the "average" change between the two points would be (f(x1)- f(x2)/(x1- x2)= 0 and the mean value theorem says that there must be a point between x1 and x2 where the derivative is 0. f'(x)= 2- cos(x). Where is that 0?
 
Uh, the "Mean Value Theorem" says that if right now you're 55 miles from your baby, and an hour ago you were 25 miles from your baby, then there must have been at least one moment when you were moving towards your baby at a rate of exactly 30 miles per hour.

It's a little more subtle than the intermediate value theorem.

Carl
 
Uh, the "Mean Value Theorem" says that if right now you're 55 miles from your baby, and an hour ago you were 25 miles from your baby, then there must have been at least one moment when you were moving towards your baby at a rate of exactly 30 miles per hour.

It's a little more subtle than the intermediate value theorem.

Carl
 
Thanks HallsofIvy and CarlB

Just one other question right now.

HallsofIvy said
If x= 0 f(x)= 2x-1-sin x= -1<0. If x=Pi, f(x)= 2x- 1- sin x= > 0

How do you know which values to use Itermediate Value? the zero and PI that you picked. Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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