91Ó°ÊÓ

We need to show that the function f(x) is continuous for all x in the domain of the function, in this case \(\mathbb{R}\). To do this, we must show that the function f(x) is continuous on the positive and negative numbers and that there is continuity at the point where the definition of the function changes (i.e. at x=0). On positive x-values, the function is defined by \(f(x)=\frac{x}{1+x}\). This is a ratio of two polynomials where the denominator is never zero. Polynomials are continuous everywhere, and a ratio of two polynomials is also continuous, provided the denominator is never zero. Hence, f(x) is continuous for \(x \geq 0\). Similarly, on negative x-values, the function is defined by \(f(x)=\frac{x}{1-x}\). This is again a ratio of two polynomials where the denominator is never zero, and thus f(x) is continuous for \(x < 0\). Now we check whether the function is continuous at x=0. The left-hand and right-hand limits must be the same: $$ \lim_{x \rightarrow 0^-} f(x) = \lim_{x \rightarrow 0^-} \frac{x}{1-x} = 0 $$ $$ \lim_{x \rightarrow 0^+} f(x) = \lim_{x \rightarrow 0^+} \frac{x}{1+x} = 0 $$ Since both one-sided limits exist and are equal, the function is continuous at x = 0. Thus, f(x) is continuous on \(\mathbb{R}\).

We have to show that f(x) is bounded on the real numbers. Observe that: For \(x \geq 0\), the function f(x) can be written as, \(f(x)=\frac{x}{1+x}\). Now, $$ 0 \leq \frac{x}{1+x} \leq 1, $$ since x is nonnegative and the denominator grows larger as x grows. For \(x < 0\), the function f(x) can be written as, \(f(x)=\frac{x}{1-x}\). Now, $$ -1 \leq \frac{x}{1-x} \leq 0, $$ since x is negative and the denominator grows smaller as x grows more negative. As a consequence, we can conclude that f(x) is bounded with: $$ -1 \leq f(x) \leq 1 $$

From the above step, we can say that the lower bound is -1 and the upper bound is 1. Now we have to show that these are indeed the infimum and the supremum. Let \(L(x)=\inf\\{f(x) : x \in \mathbb{R}\\}\) and \(U(x)=\sup\\{f(x) : x \in \mathbb{R}\\}\). If \(0 \leq x < 1\), then \(0 < 1 - x \leq 1\), so \(-1 \leq \frac{x}{1-x} < 0\). Thus, \(L(x)=-1\). If \(0 \leq x < 1\), then \(1 \leq 1 + x\), so \(0 < \frac{x}{1+x} < 1\). Thus, \(U(x)=1\).

Now we have to show that there are no values r and s in \(\mathbb{R}\) such that \(f(r)=-1\) and \(f(s)=1\). Let's assume that there exists an r such that f(r)=-1. For \(r < 0\), we have: $$ f(r) = \frac{r}{1-r} = -1 $$ Multiplying both sides by (1-r), $$ r = -(1 - r) $$ Adding r to both sides, $$ 2r = 1, $$ which implies \(r = \frac{1}{2}\). But this value of r contradicts the fact that r must be negative; therefore, no such r exists. Now, let's assume there exists an s such that f(s)=1. For \(s \geq 0\), we have: $$ f(s) = \frac{s}{1+s} = 1 $$ Multiplying both sides by (1+s), $$ s = 1 + s $$ Subtracting s from both sides, $$ 0 = 1, $$ which is a contradiction. Thus, no such s exists. Hence, we have demonstrated that \(f(x)\) is continuous and bounded on \(\mathbb{R}\), that \(\inf\\{f(x):x \in \mathbb{R}\\}=-1\), \(\sup\\{f(x):x \in \mathbb{R}\\}=1\), and that there do not exist \(r, s\) in \(\mathbb{R}\) such that \(f(r)=-1\) and \(f(s)=1\).