tricky Continuity

[squaremeplz]

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.

[HallsofIvy]

Why not just the basic definition of continuity?

If f is continuous at c, then, given any \epsilon> 0, there exist \delta>0 such that if |x- c|< \delta, then [/itex]|f(x)- f(c)|< \epsilon[/itex].

Take \epsilon= 1/2.

[squaremeplz]

was part 1. correct? I still feel a bit insecure about my answer. thanks

[HallsofIvy]

Yes, part 1 is correct.

[Dick]

was part 1. correct? I still feel a bit insecure about my answer. thanks

Your proof of part 1 is a little confusing. Consider the function g(x)=x-cos(x)^2. Can you show g(x) has a zero in [0,pi/2]?