1. The problem statement, all variables and given/known data 1. show there is some point x in the interval [0,pi/2] so that x = cos(x)^2 2. let f:R-> be continuous at c and suppose f(c) =1. show that there is some a > 0 such that f(x) > 1/2 whenever |x-c| < a 2. Relevant equations intermediate value theorem. maximum minum 3. The attempt at a solution 1. We know that f(x) = cos(x)^2 assumes its max value at f(0) = 1 and its min. at f(pi/2) = 0 by the int. value theoreom 0 < pi/2 implies f(pi/2) < y < f(0) implies for some z in [0,pi/2] f(z) = y the only conclusion I can come up with is that since the interval [0,pi/2] is greater than the distance between |f(0)-f(pi/2)| there has to be at least one x in [0,pi/2] where f(x) = x. Is this plausible? b) Unsure as to how to properly approach this problem. since f(x) > 1/2 |f(x) - f(c)| = |1/2 - 1| <= |1/2| < e/2 let a equal e/2? I truly apologize for this poor approach. If someone could just direct me in the right direction it would be greatly appreciated.