Solve Tricky Continuity Homework Statement

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In summary, for the first problem, we can use the intermediate value theorem to show that there exists a point x in the interval [0,pi/2] where x = cos(x)^2. This is because the function f(x) = cos(x)^2 has a maximum value of 1 at x = 0 and a minimum value of 0 at x = pi/2, and by the intermediate value theorem, there must be a point in between where f(x) takes on any value in between. For the second problem, we can use the definition of continuity to show that there exists an a > 0 such that f(x) > 1/2 whenever |x-c| < a.
  • #1
squaremeplz
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Homework Statement



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

Homework Equations



intermediate value theorem. maximum minum

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.
 
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  • #2
Why not just the basic definition of continuity?

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

Take [itex]\epsilon= 1/2[/itex].
 
  • #3
was part 1. correct? I still feel a bit insecure about my answer. thanks
 
  • #4
Yes, part 1 is correct.
 
  • #5
squaremeplease said:
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]?
 

What is continuity in mathematics?

Continuity is a mathematical concept that describes the smoothness of a function or a curve. It means that there are no gaps, jumps, or sudden changes in the values of the function as the input values change.

How do you determine if a function is continuous?

A function is considered continuous if it satisfies three conditions: the function is defined at the point in question, the limit of the function at that point exists, and the limit is equal to the value of the function at that point.

What is a removable discontinuity?

A removable discontinuity, also known as a hole, is a type of discontinuity where a function has a gap at a particular point, but this gap can be filled by assigning a value to that point. In other words, the limit of the function at that point exists, but the function is not defined at that point.

What is a jump discontinuity?

A jump discontinuity is a type of discontinuity where there is a sudden change in the values of a function at a particular point. This means that the limit of the function at that point does not exist.

How do you solve tricky continuity homework statements?

Solving tricky continuity homework statements requires a thorough understanding of the concept of continuity and knowing the different types of discontinuities. It also involves using algebraic and graphical techniques to analyze the behavior of the function at a particular point. It is important to carefully examine the conditions of continuity and to consider special cases such as removable discontinuities and jump discontinuities.

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