Given that \( f(x)=(x-6)^{2} \) and \( g(x)=2-3 x \), find (a) \( (f+g)(x)=\square \) (b) \( (f-g)(x)=\square \) (c) \( (f g)(x)=\square \) (d) \( \left(\frac{f}{g}\right)(x)=\square \) Question Help:

Let's tackle the given functions step by step! (a) \( (f+g)(x) = f(x) + g(x) = (x-6)^{2} + (2 - 3x) \) To expand \( (x-6)^{2} \): \[ (x-6)^{2} = x^{2} - 12x + 36 \] Now, add \( (2 - 3x) \): \[ (f+g)(x) = x^{2} - 12x + 36 + 2 - 3x = x^{2} - 15x + 38 \] (b) \( (f-g)(x) = f(x) - g(x) = (x-6)^{2} - (2 - 3x) \) Using the expansion from above: \[ (f-g)(x) = (x^{2} - 12x + 36) - (2 - 3x) \] \[ = x^{2} - 12x + 36 - 2 + 3x = x^{2} - 9x + 34 \] (c) \( (f g)(x) = f(x) \cdot g(x) = (x-6)^{2} \cdot (2 - 3x) \) First, we can expand \( (x-6)^{2} \): \[ = (x^{2} - 12x + 36)(2 - 3x) \] To simplify, \[ = 2(x^{2} - 12x + 36) - 3x(x^{2} - 12x + 36) \] \[ = 2x^{2} - 24x + 72 - 3x^{3} + 36x - 108 \] Collecting like terms yields: \[ (f g)(x) = -3x^{3} + 2x^{2} + 12x - 36 \] (d) \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{(x-6)^{2}}{2 - 3x} \) This is already in a simplified form, but you can rewrite it as: \[ \left(\frac{f}{g}\right)(x) = \frac{x^{2} - 12x + 36}{2 - 3x} \] Just be cautious with the domain: \( g(x) \neq 0 \Rightarrow 2 - 3x \neq 0 \) which gives \( x \neq \frac{2}{3} \). That's it! You now have \( (f+g)(x) \), \( (f-g)(x) \), \( (f g)(x) \), and \( \left(\frac{f}{g}\right)(x) \) elegantly expressed!