1. Let \( f \) be a function that is everywhere differentiable and that has the following properties. (i) \( f(x+h)=\frac{f(x)+f(h)}{f(-x)+f(-h)} \) for all real numbers \( h \) and \( x \). (ii) \( f(x)>0 \) for all real numbers \( x \). (iii) \( f^{\prime}(0)=-1 \). (a) Find the value of \( f(0) \). (b) Show that \( f(-x)=\frac{1}{f(x)} \) for all real numbers \( x \). (c) Using part (b), show that \( f(x+h)=f(x) f(h) \) for all real numbers \( h \) and \( x \). (d) Use the definition of the derivative to find \( f(x) \) in terms of \( f(x) \).

To solve the problem step by step, we will analyze each part carefully. ### Part (a): Find the value of \( f(0) \). We start with the functional equation given in part (i): \[ f(x+h) = \frac{f(x) + f(h)}{f(-x) + f(-h)} \] Let's substitute \( x = 0 \) into this equation: \[ f(0+h) = f(h) = \frac{f(0) + f(h)}{f(0) + f(-h)} \] This simplifies to: \[ f(h) (f(0) + f(-h)) = f(0) + f(h) \] Rearranging gives: \[ f(h) f(0) + f(h) f(-h) = f(0) + f(h) \] Now, we can isolate terms involving \( f(h) \): \[ f(h) f(0) + f(h) f(-h) - f(h) = f(0) \] Factoring out \( f(h) \): \[ f(h) (f(0) + f(-h) - 1) = f(0) \] Since \( f(h) > 0 \) for all \( h \), we can divide both sides by \( f(h) \): \[ f(0) + f(-h) - 1 = \frac{f(0)}{f(h)} \] Now, let’s evaluate this at \( h = 0 \): \[ f(0) + f(0) - 1 = \frac{f(0)}{f(0)} \] This simplifies to: \[ 2f(0) - 1 = 1 \] Thus, we have: \[ 2f(0) = 2 \implies f(0) = 1 \] ### Part (b): Show that \( f(-x) = \frac{1}{f(x)} \) for all real numbers \( x \). Using the functional equation again, we substitute \( h = 0 \): \[ f(x + 0) = f(x) = \frac{f(x) + f(0)}{f(-x) + f(0)} \] Substituting \( f(0) = 1 \): \[ f(x) = \frac{f(x) + 1}{f(-x) + 1} \] Cross-multiplying gives: \[ f(x)(f(-x) + 1) = f(x) + 1 \] Expanding and rearranging: \[ f(x) f(-x) + f(x) = f(x) + 1 \] This simplifies to: \[ f(x) f(-x) = 1 \] Thus, we conclude: \[ f(-x) = \frac{1}{f(x)} \] ### Part (c): Show that \( f(x+h) = f(x) f(h) \) for all real numbers \( h \) and \( x \). Using the result from part (b): \[ f(x+h) = \frac{f(x) + f(h)}{f(-x) + f(-h)} \] Substituting \( f(-x) = \frac{1}{f(x)} \) and \( f(-h) = \frac{1}{f(h)} \): \[ f(x+h) = \frac{f(x) + f(h)}{\frac{1}{f(x)} + \frac{1}{f(h)}} \] This simplifies to: \[ f(x+h) = \frac{f(x) + f(h)}{\frac{f(h) + f(x)}{f(x) f(h)}} \] Thus: \[ f(x+h) = f(x) f(h) \] ### Part (d): Use the definition of the derivative to find \( f(x) \) in terms of \( f(x) \). We know that \( f(x+h) = f(x) f(h) \). Taking the derivative with respect to \( x \): \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{f(x) f(h) - f(x)}{h} \] Factoring out \( f(x) \): \[ f'(x) = f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \] Let \( L = \lim_{h \to 0} \frac{f(h) - 1}{h} \). We know from part (iii) that \( f'(0) = -1 \), so: \[ L = f'(0) = -1 \] Thus: \[ f'(x) = -f(x) \] This is a first-order linear differential equation. The general solution is: \[ f(x) = Ce^{-x} \] Using the condition \( f(0) = 1 \): \[ f(0) = C \cdot e^0 = C = 1 \] Thus, we have: \[ f(x) = e^{-x} \] ### Summary of Results: - (a) \( f(0) = 1 \) - (b) \( f(-x) = \frac{1}{f(x)} \) - (c) \( f(x+h) = f(x) f(h) \) - (d) \( f(x) = e^{-x} \)