Question

This question is meant to introduce a result which should have been covered in an standard introductory calculus course (with or without proof), which has wide application, whose validity looks heuristically obvious but whose proof is nontrivial, and which, from the historical point of view, motivated mathematicians to search for a rigorous foundation for calculus. We introduce the following result in calculus, known as Bolzano's Intermediate Value Theorem (BIVT):

(BIVT) Let a,b∈R, with a In this question, you may take for granted the validity of this result, alongside that of other basic results concerned with limit and continuity covered in your calculus courses, and use these results where appropriate and necessary. Let f:[0,1]→R be a function. Suppose f is continuous on [0,1], and x² +(f(x))² =1 for any x∈[0,1].
(a) Suppose p∈[0,1]. Prove that f(p)=0 iff p=1.
(b) Suppose f(0)≥0.
i. What is the value of f(0) ?
ii. Prove that f(x)= √1−x ²for any x∈[0,1].
Remark. Bolzano's Intermediate Value Theorem may become relevant at some stage in the argument.
(c)Suppose f(0)≤0. Prove that f(x)=−√1−x ² for any x∈[0,1].

Answer 1

We have a function f(x) = √(1-x²), and we need to prove several statements based on its properties. (a) We prove that f(p) = 0 implies p = 1. (b) We find the value and expression of f(x) when f(0)≥0. (c) We find the value and expression of f(x) when f(0)≤0.

Step-by-step explanation:

In this question, we are given a function f(x) = √(1-x²) and asked to prove certain statements based on its properties. Let's go through each part:

(a) Suppose p∈[0,1]. To prove f(p)=0 if and only if p=1, we can start by assuming f(p)=0 and then show that this implies p=1. Since f(p)=0, it means √(1-p²)=0, which implies 1-p²=0 and therefore p²=1. Taking the square root of both sides, we get p = ±1. However, since p is in the interval [0, 1], only p = 1 is valid, so f(p) = 0 implies p = 1.

(b) Suppose f(0)≥0. (i) To find the value of f(0), we can substitute x=0 into the given function, which gives us f(0) = √(1-0²) = 1. (ii) To prove that f(x)=√(1-x²) for any x∈[0,1], we can start by assuming arbitrary x in this interval and then show that f(x)=√(1-x²) holds. Substituting x into the function, we have f(x) = √(1-x²), which matches the desired expression.

(c) Suppose f(0)≤0. To prove that f(x)=-√(1-x²) for any x∈[0,1], we can follow a similar approach as in part (b). By substituting x into the function, we have f(x) = -√(1-x²), which matches the desired expression.