Determine if the following functions have crossover points: \( \begin{array}{ll}\text { a) } f(x)=\frac{(2-x)(3 x+2)}{x^{2}} & \text { b) } f(x)=\frac{2 x^{3}-3 x^{2}+x-3}{x^{2}+1}\end{array} \)

To determine if the given functions have crossover points (i.e., points where \( f(x) = 0 \)), we'll analyze each function separately. ### a) \( f(x) = \frac{(2 - x)(3x + 2)}{x^2} \) **Finding Crossover Points:** Set \( f(x) = 0 \): \[ \frac{(2 - x)(3x + 2)}{x^2} = 0 \] The fraction equals zero when the **numerator** is zero (since the denominator \( x^2 \) cannot be zero). \[ (2 - x)(3x + 2) = 0 \] Solving for \( x \): 1. \( 2 - x = 0 \) \( x = 2 \) 2. \( 3x + 2 = 0 \) \( x = -\frac{2}{3} \) **Conclusion for Part a):** The function \( f(x) \) has two crossover points at \( x = 2 \) and \( x = -\frac{2}{3} \). --- ### b) \( f(x) = \frac{2x^3 - 3x^2 + x - 3}{x^2 + 1} \) **Finding Crossover Points:** Set \( f(x) = 0 \): \[ \frac{2x^3 - 3x^2 + x - 3}{x^2 + 1} = 0 \] Again, the fraction equals zero when the **numerator** is zero. \[ 2x^3 - 3x^2 + x - 3 = 0 \] **Solving the Cubic Equation:** - **Rational Root Theorem:** Possible rational roots are \( \pm1, \pm3, \pm\frac{1}{2}, \pm\frac{3}{2} \). - **Testing Possible Roots:** - \( x = 1 \): \( 2(1)^3 - 3(1)^2 + 1 - 3 = 2 - 3 + 1 - 3 = -3 \neq 0 \) - \( x = -1 \): \( 2(-1)^3 - 3(-1)^2 + (-1) - 3 = -2 - 3 - 1 - 3 = -9 \neq 0 \) - \( x = \frac{3}{2} \): \( 2\left(\frac{27}{8}\right) - 3\left(\frac{9}{4}\right) + \frac{3}{2} - 3 = \frac{27}{4} - \frac{27}{4} + \frac{3}{2} - 3 = -\frac{3}{2} \neq 0 \) Since none of the rational roots work, we'll analyze the behavior of the cubic polynomial: - **End Behavior:** - As \( x \to \infty \), \( 2x^3 \) dominates, so \( f(x) \to \infty \). - As \( x \to -\infty \), \( 2x^3 \) dominates, so \( f(x) \to -\infty \). - **Derivative for Critical Points:** - \( f'(x) = 6x^2 - 6x + 1 \) - Solving \( f'(x) = 0 \) gives two critical points, indicating potential changes in direction. Given the behavior and absence of additional real roots from rational testing, the cubic equation likely has **one real root**. **Conclusion for Part b):** The function \( f(x) \) has **one** crossover point. --- ### **Final Answer:** a) Two crossover points; b) One crossover point.