# How to solve this function math problem?

Find the maximum value and the minimum value of the following function. Also give the least possible values of x at which they occur.

30/ 11-5cos(1/2x-45)

I'm gonna assume that your functions takes x as degrees.

f(x) = 30/11 - 5Cos( (1/2)x - 45)

First of all, the maximum value of all cosine functions is equal to 1and the minimum value is always -1. But since f(x) uses a negative cosine, the maximum value of the function occurs during the minimum value of cosine.

Therefore, let because( (1/2)x - 45) = -1 Max Value = (30/11) - 5(-1) = (30/11) + 5 = (30/11) + (55/11) = 85/11

And the minimum value of the function occurs at the maximum value of cosine. Therefore set because( (1/2)x - 45) = 1

Min. Value = (30/11) - 5(1) = (30/11) - 5 = (30/11) - (55/11) = -25/11

To find the x values for which these minima and maxima occur, remember that the min. value of f(x) occurs at the max. value of cosine. Solving the equation because( (1/2)x - 45) = 1 for x will give you the value of x at which f(x) is minimum and solving because ( (1/2)x - 45) = -1 for x will give you the value of x at which f(x) is maximum.

Min. Max.

because( (1/2)x - 45) = 1 because ( (1/2)x - 45) = -1 (1/2)x - 45 = 0 (1/2)x - 45 = 180 (1/2)x = 45 (1/2)x = 225 x = 90 x = 450

Therefore minimum value of f(x) occurs at x = 90 degrees and maximum value of f(x) occurs at x = 450 degrees

It's too early for me to try, so here's a link to a math search engine with your problem:

http://www.wolframalpha.com/input/?I=30%2F+11-5cos%281%2F2x-45%29

still not getting the answer I want(I.e the one written at the back of my book) lol =p

but thanks again 4 the link...its helping me with other problems :)