Prove Inequality using Mean Value Theorem

In summary, the conversation discusses using the mean value theorem (MVT) to prove the inequality abs(sina - sinb) \leq abs(a - b) for all a and b. The person asking the question is struggling with the concept and is unsure how to approach it without a graphing calculator. The other person suggests using the MVT and taking the absolute value of both sides, using the fact that |cos(c)|<=1. It is also mentioned that the absolute value of sin will look like a sequence of upside-down cups with vertical tangents between them.
  • #1
Khayyam89
7
0

Homework Statement


Essentially, the question asks to use the mean value theorem(mvt) to prove the inequality: abs(sina - sinb) [tex]\leq[/tex] abs(a - b) for all a and b


The Attempt at a Solution



I do not have a graphing calculator nor can I use one for this problem, so I need to prove that the inequality basically by proof. What I did was to look at the mvt hypotheses: if the function is continuous and differetiable on closed and open on interval a,b, respectively. However, the problem I am having is that I am getting thrown off by the absolute values and the fact that I've never used mvt on inequalities. I know the absolute value of the sin will look like a sequence of upside-down cups with vertical tangents between them. Hints most appreciated.
 
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  • #2
Assume a>b, then sina-sinb=(a-b)cosc, for some b<c<a, which gives the inequality with no problems.
 
  • #3
Wait, are you considering that abs(...) = absolute value of the sum?
 
  • #4
The mean value theorem tells you (sin(a)-sin(b))/(a-b)=sin'(c)=cos(c) for some c between a and b, as boombaby said. Take the absolute value of both sides and use that |cos(c)|<=1.
 

1. What is the Mean Value Theorem?

The Mean Value Theorem is a fundamental theorem in calculus that states that for any continuous and differentiable function on an interval, there exists a point within that interval where the instantaneous rate of change (derivative) equals the average rate of change (slope of the secant line). This point is known as the "mean value" and can be used to prove various inequalities.

2. How can the Mean Value Theorem be used to prove inequalities?

The Mean Value Theorem can be used to prove inequalities by showing that the derivative of a function is either always positive or always negative on a given interval. This information can then be used to determine the behavior of the function and prove that it is either always greater than or always less than another function on that interval.

3. What are the steps for proving an inequality using the Mean Value Theorem?

The steps for proving an inequality using the Mean Value Theorem are as follows:

  1. Identify the functions involved in the inequality and the interval on which it is being considered.
  2. Show that the functions are continuous and differentiable on the given interval.
  3. Find the derivative of the function and show that it is either always positive or always negative on the interval.
  4. Use the Mean Value Theorem to find the "mean value" or point where the derivative equals the average rate of change.
  5. Compare the values of the two functions at the "mean value" point to prove the inequality.

4. Can the Mean Value Theorem be used to prove all types of inequalities?

No, the Mean Value Theorem can only be used to prove inequalities between continuous and differentiable functions on an interval. It cannot be used to prove inequalities involving discrete functions or functions that are not differentiable.

5. Are there any limitations to using the Mean Value Theorem to prove inequalities?

Yes, there are some limitations to using the Mean Value Theorem to prove inequalities. For example, if the functions involved are not continuous on the given interval, the theorem cannot be applied. Additionally, if the functions have multiple points where the derivative is equal to the average rate of change, the "mean value" point may not be unique and the inequality may not hold.

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