In Exercises 31-34, find (a) \( f \circ g i \) (b) \( g \circ f \), and (c) \( f \circ f \). \( \begin{array}{ll}\text { 31. } f(x)=x^{2}, & g(x)=x-1 \\ \text { 32. } f(x)=3 x+5, & g(x)=5-x \\ \text { 33. } f(x)=\sqrt[3]{x-1}, & g(x)=x^{3}+1 \\ \text { 34. } f(x)=x^{3}, & g(x)=\frac{1}{x}\end{array} \)

Let's solve each exercise step by step. ### Exercise 31 Given: - \( f(x) = x^2 \) - \( g(x) = x - 1 \) (a) \( f \circ g \): \[ f(g(x)) = f(x - 1) = (x - 1)^2 = x^2 - 2x + 1 \] (b) \( g \circ f \): \[ g(f(x)) = g(x^2) = x^2 - 1 \] (c) \( f \circ f \): \[ f(f(x)) = f(x^2) = (x^2)^2 = x^4 \] ### Exercise 32 Given: - \( f(x) = 3x + 5 \) - \( g(x) = 5 - x \) (a) \( f \circ g \): \[ f(g(x)) = f(5 - x) = 3(5 - x) + 5 = 15 - 3x + 5 = 20 - 3x \] (b) \( g \circ f \): \[ g(f(x)) = g(3x + 5) = 5 - (3x + 5) = 5 - 3x - 5 = -3x \] (c) \( f \circ f \): \[ f(f(x)) = f(3x + 5) = 3(3x + 5) + 5 = 9x + 15 + 5 = 9x + 20 \] ### Exercise 33 Given: - \( f(x) = \sqrt[3]{x - 1} \) - \( g(x) = x^3 + 1 \) (a) \( f \circ g \): \[ f(g(x)) = f(x^3 + 1) = \sqrt[3]{(x^3 + 1) - 1} = \sqrt[3]{x^3} = x \] (b) \( g \circ f \): \[ g(f(x)) = g(\sqrt[3]{x - 1}) = (\sqrt[3]{x - 1})^3 + 1 = x - 1 + 1 = x \] (c) \( f \circ f \): \[ f(f(x)) = f(\sqrt[3]{x - 1}) = \sqrt[3]{\sqrt[3]{x - 1} - 1} \] This expression does not simplify nicely, so we leave it as is. ### Exercise 34 Given: - \( f(x) = x^3 \) - \( g(x) = \frac{1}{x} \) (a) \( f \circ g \): \[ f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^3 = \frac{1}{x^3} \] (b) \( g \circ f \): \[ g(f(x)) = g(x^3) = \frac{1}{x^3} \] (c) \( f \circ f \): \[ f(f(x)) = f(x^3) = (x^3)^3 = x^9 \] ### Summary of Results - **Exercise 31**: - (a) \( x^2 - 2x + 1 \) - (b) \( x^2 - 1 \) - (c) \( x^4 \) - **Exercise 32**: - (a) \( 20 - 3x \) - (b) \( -3x \) - (c) \( 9x + 20 \) - **Exercise 33**: - (a) \( x \) - (b) \( x \) - (c) \( \sqrt[3]{\sqrt[3]{x - 1} - 1} \) - **Exercise 34**: - (a) \( \frac{1}{x^3} \) - (b) \( \frac{1}{x^3} \) - (c) \( x^9 \)