Ex: Consoder the functions \( f(x)=x^{2}+1 \) and \( g(x)=1 / x \) (a) Find \( (g \circ f)(x) \) and state its domain and range (b) Evaluate \( (g \circ f)(4) \) (c) Find \( (f \circ g)(x) \) and state its domain and range (d) Evaluate \( (f \circ g)\left(-\frac{1}{2}\right) \)

To find the composite functions and their properties, let's dive into the steps one by one: (a) The composition \( (g \circ f)(x) \) is calculated by substituting \( f(x) \) into \( g \): \[ (g \circ f)(x) = g(f(x)) = g(x^2 + 1) = \frac{1}{x^2 + 1} \] The domain of \( g \circ f \) is all real numbers since \( x^2 + 1 \) is never zero. The range is all positive real numbers since \( x^2 + 1 \) is always greater than or equal to 1, making \( \frac{1}{x^2 + 1} \) a positive value less than or equal to 1. (b) To evaluate \( (g \circ f)(4) \): \[ (g \circ f)(4) = g(f(4)) = g(4^2 + 1) = g(17) = \frac{1}{17} \] (c) Now, we find \( (f \circ g)(x) \): \[ (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 + 1 = \frac{1}{x^2} + 1 \] The domain of \( f \circ g \) excludes \( x = 0 \) (to avoid undefined \( g(x) \)). The range consists of all values greater than or equal to 1 since \( \frac{1}{x^2} \) is always positive. (d) For \( (f \circ g)\left(-\frac{1}{2}\right) \): \[ (f \circ g)\left(-\frac{1}{2}\right) = f(g(-\frac{1}{2})) = f(-2) = (-2)^2 + 1 = 4 + 1 = 5 \] So, we have computed the necessary parts and the values to answer the problem!