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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)
please please help! I got my maths exam tomorrow!!
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 :)
Hmm. Try it this way (substituted .5x for 1/2x)
http://www.wolframalpha.com/input/?I=30+%2F+11+-5cos%28.5x-45%29
thx for the link! Bur unfortunately it didn't give me the solution I required