Absolute Extrema on Closed Interval for f(x) = x-2cosx

In summary, the task at hand is to find the absolute maximum and absolute minimum values of the function f(x) = x-2cosx on the given interval [-pi, pi]. To do so, the first step is to calculate the derivative of f and find its zeroes, which will be the critical values of f. Then, calculate the value of f at these critical points, as well as at the endpoints of the interval (-pi and pi). Finally, compare all the values of f calculated to determine the absolute maximum and minimum values. In this case, the absolute minimum is found at f(-5pi/6) = -5pi/6 - root3, while the absolute maximum is found at f(pi) = (pi
  • #1
loadsy
57
0
The question I'm currently working on now is to find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x-2cosx, [-pi, pi]
f'(x) = 1+2sinx = 0

From here I've found that -30 degrees(-pi/6) and -150(-5pi/6) degrees are the two intercepts in the equation. However I'm just having some trouble figuring out how to solve for this type of question. Any sort of hints would be great, thanks :P
 
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  • #2
Inside the open interval (-pi,pi), you can calculate the derivative of f. The zeroes of the derivative are the critical values of f (either a min, a max or a saddle point). Calculate the value of f at each of these points. Also calculate f directly at -pi and pi. Now compare all the values of f you've calculated. The bigger is the absolute max and the smallest is the absolute min.
 
  • #3
Ahhh alrighty that makes sense. So pretty much just find the derivatives of f(-pi/6) and f(-5pi/6) along with f(-pi) and f(pi). Because I'm finding that the derivative of -pi/6 and -5pi/6 are 0 thus they are both critical numbers of f on (-pi,pi).
 
  • #4
Okay boo yeah, I think I figured it out, f(-5pi/6) = -5pi/6 - root3 is the absolute minimum and the absolute maximum is f(pi) = (pi) + 2
 

Related to Absolute Extrema on Closed Interval for f(x) = x-2cosx

1. What is the Method of Closed Interval?

The Method of Closed Interval is a numerical method used to approximate the roots of a function within a given interval. It is also known as the Bisection Method or the Interval Halving Method.

2. How does the Method of Closed Interval work?

The Method of Closed Interval works by repeatedly dividing the given interval in half and checking for a sign change in the function. If there is a sign change, the root is assumed to be within that interval and the process is repeated until the desired level of accuracy is achieved.

3. What are the advantages of using the Method of Closed Interval?

The Method of Closed Interval is relatively easy to implement and can provide accurate results for a wide range of functions. It also guarantees convergence to a root if the function is continuous and there is a sign change within the given interval.

4. Are there any limitations of the Method of Closed Interval?

One limitation of the Method of Closed Interval is that it may take a large number of iterations to achieve the desired level of accuracy, especially for functions with multiple roots or when the root is close to the boundaries of the interval. It also does not work well for functions that do not have a sign change within the interval.

5. How is the Method of Closed Interval different from other root finding methods?

The Method of Closed Interval is a bracketing method, meaning it requires an initial interval where the root is located. This makes it more reliable compared to open methods, such as the Newton-Raphson method, which can fail to converge if the initial guess is not close enough to the root. However, bracketing methods may require more computation and can be slower compared to open methods.

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