Question

Determine whether the following statements are true and give an explanation or counterexample. a. If a function is left-continuous and right-continuous at \(a\), then it is continuous at \(a\) b. If a function is continuous at \(a\), then it is left-continuous and right- continuous at \(a\) c. If \(a< b\) and \(f(a) \leq L \leq f(b),\) then there is some value of \(c\) in \((a, b)\) for which \(f(c)=L\) d. Suppose \(f\) is continuous on \([a, b] .\) Then there is a point \(c\) in \((a, b)\) such that \(f(c)=(f(a)+f(b)) / 2\)

Short answer

Question: Determine the truth or falsehood of the following statements, providing explanations or counterexamples as necessary. a. If a function is left-continuous and right-continuous at \(a\), then it is continuous at \(a\). b. If a function is continuous at \(a\), then it is left-continuous and right-continuous at \(a\). c. If \(a< b\) and \(f(a) \leq L \leq f(b),\) there is some value of \(c\) in \((a, b)\) for which \(f(c)=L\). d. Suppose \(f\) is continuous on \([a, b]\). Then there is a point \(c\) in \((a,b)\) such that \(f(c)=(f(a)+f(b))/2\). Answer: a. True. Left-continuous and right-continuous functions at a point \(a\) satisfy the definition of continuity at that point. b. True. A continuous function at a point \(a\) satisfies the definitions of both left-continuity and right-continuity at that point. c. False. A counterexample is provided with a function that is not continuous at \(x=0\). It satisfies \(f(a) < L < f(b)\) for chosen \(a, b, L\), but there is no \(c\) in the interval \((-1, 1)\) such that \(f(c) = 0\). d. True. The statement follows from the Intermediate Value Theorem for the specific value of \(L=(f(a)+f(b))/2\).

Step-by-step solution

Statement a: Analyzing left-continuous and right-continuous functions

To prove that this statement is true, we must show that left-continuous and right-continuous functions at a point \(a\) are continuous at that point. They are defined as follows: 1. A function \(f(x)\) is left-continuous at \(a\) if \(\lim_{x \to a-}f(x) = f(a)\). That is, the function approaches its value as \(x\) approaches \(a\) from the left. 2. A function \(f(x)\) is right-continuous at \(a\) if \(\lim_{x \to a+}f(x) = f(a)\). That is, the function approaches its value as \(x\) approaches \(a\) from the right. If both of these conditions are met, then the limit of the function as \(x\) approaches \(a\) is equal to the value of the function at \(a\), which is the definition of continuity.

Statement a: Conclusion

The statement is true because a function that is both left-continuous and right-continuous at a point \(a\) satisfies the definition of continuity at that point.

Statement b: Analyzing continuous functions

Continuity at a point \(a\) is defined as \(\lim_{x \to a}f(x) = f(a)\). We previously established that left-continuity and right-continuity imply continuity. Since the limit from both left and right must exist and be equal for a function to be continuous, this implies left-continuity and right-continuity as well.

Statement b: Conclusion

The statement is true because a continuous function at a point \(a\) satisfies the definitions of both left-continuity and right-continuity at that point.

Statement c: Analyzing the Intermediate Value Theorem

This statement is related to the Intermediate Value Theorem (IVT). The IVT states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\), and if \(L\) is any value between \(f(a)\) and \(f(b)\), then there exists a number \(c\) in the open interval \((a, b)\) such that \(f(c) = L\). In our statement, the requirement for the function to be continuous is missing. Let's see if we can find a counterexample to prove that the statement is false.

Statement c: Counterexample

Consider the function \(f(x) = x\) for \(x\leq 0\) and \(f(x) = x+1\) for \(x > 0\). This function is not continuous at \(x=0\). Let \(a=-1\), \(b=1\), and \(L=0\). We have \(f(a)=-1\), \(f(b)=2\), and \(f(a) < L < f(b)\). However, there is no \(c\) in the interval \((-1, 1)\) such that \(f(c) = 0\).

Statement c: Conclusion

The statement is false, as demonstrated by the provided counterexample.

Statement d: Analyzing continuous functions on closed intervals

This statement is related to the Mean Value Theorem, which states that if a function \(f(x)\) is continuous on a closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a point \(c\) in the open interval \((a, b)\) such that \(f'(c) = (f(b)-f(a))/(b-a)\). However, our statement does not require differentiability, and we are to find a point \(c\) such that \(f(c) = (f(a)+f(b))/2\). This is basically an application of the Intermediate Value Theorem (IVT) but with a specific value of \(L\).

Statement d: Explanation

Using the IVT, we know that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and if \(L\) is any value between \(f(a)\) and \(f(b)\), then there exists a number \(c\) in the open interval \((a, b)\) such that \(f(c) = L\). In this case, we have \(L = (f(a)+f(b))/2\). Since \(f\) is continuous on \([a, b]\), by the IVT, there must exist a point \(c\) in \((a, b)\) such that \(f(c) = (f(a)+f(b))/2\).

Statement d: Conclusion

The statement is true, based on the application of the Intermediate Value Theorem for the specific value of \(L=(f(a)+f(b))/2\).

Key concepts

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) is a cornerstone of calculus that captures the intuitive idea that a continuous function, one without jumps or breaks, necessarily passes through every value between any two points on the function's graph.

To put it formally, the IVT states that if a function f(x) is continuous on a closed interval \[a, b\], and L is any number between f(a) and f(b), there is at least one number c in the open interval \(a, b\) such that f(c) = L. In the context of the given exercise, this theorem justifies the existence of a point c where the function reaches the average value of its endpoints, (f(a)+f(b))/2, assuming continuity over the interval.

Left-Continuity

Left-continuity pertains to how a function behaves as you approach a specific point a from the left. When we say a function f(x) is left-continuous at point a, we mean that as x gets closer and closer to a from values less than a, the function's value approaches f(a) itself. This can be expressed using limits: \[\lim_{x \to a-}f(x) = f(a)\].

The idea is that there's no sudden jump or gap in the function as you move from left to right towards the point a. In practice, if you were to draw the function's graph, you would be able to draw right up to f(a) without lifting your pencil.

Right-Continuity

Right-continuity mirrors left-continuity but focuses on the behavior of function as you approach point a from the right. A function f(x) is said to be right-continuous at a if, when approaching a from values greater than a, the function's value gets closer and closer to f(a). Symbolically, we have: \[\lim_{x \to a+}f(x) = f(a)\].

This is to say, if you were moving along the graph from right to left, as you reach point a, there would be no gap or sudden leap in the function’s value. Right-continuity ensures that the function’s value at a is in harmony with the values immediately to its right.

Mean Value Theorem

The Mean Value Theorem (MVT) is another fundamental component of calculus, which relates the average rate of change of a function to the instantaneous rate of change at a particular point. The formal statement is that if a function f(x) is continuous on a closed interval \[a, b\] and differentiable on the open interval \(a, b\), then there exists at least one point c in \(a, b\) where \[f'(c) = \frac{f(b)-f(a)}{b-a}\].

The theorem provides a bridge between the average change over an interval and the function's derivative, which represents the instantaneous change at a particular point. MVT ensures that somewhere between a and b, the function's slope is exactly equal to the average slope between these endpoints.