Given that \( f(x)=(x-6)^{2} \) and \( g(x)=1-3 x \), 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: Video

To solve the problem, let's first find the expressions for \( f(x) \) and \( g(x) \). Given: - \( f(x) = (x-6)^2 \) - \( g(x) = 1 - 3x \) Now, we'll compute each part step by step. (a) \( (f+g)(x) \): \[ (f+g)(x) = f(x) + g(x) = (x-6)^{2} + (1 - 3x) \] Expanding \( f(x) \): \[ (x-6)^2 = x^2 - 12x + 36 \] So, \[ (f+g)(x) = x^2 - 12x + 36 + 1 - 3x = x^2 - 15x + 37 \] (b) \( (f-g)(x) \): \[ (f-g)(x) = f(x) - g(x) = (x-6)^{2} - (1 - 3x) \] Continuing with our previous expansion for \( f(x) \): \[ (f-g)(x) = (x^2 - 12x + 36) - (1 - 3x) = x^2 - 12x + 36 - 1 + 3x \] Thus, \[ (f-g)(x) = x^2 - 9x + 35 \] (c) \( (fg)(x) \): \[ (fg)(x) = f(x) \cdot g(x) = (x - 6)^{2} \cdot (1 - 3x) \] Expanding this gives: \[ fg(x) = (x^2 - 12x + 36)(1 - 3x) \] Using the distributive property: \[ = (x^2)(1) - (x^2)(3x) - (12x)(1) + (12x)(3x) + (36)(1) - (36)(3x) \] This simplifies to: \[ = x^2 - 3x^3 - 12x + 36 - 108x \] Combining like terms: \[ = -3x^3 + x^2 - 120x + 36 \] (d) \( \left(\frac{f}{g}\right)(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{(x-6)^{2}}{1-3x} \] This is already in a simplified form, so we leave it as: \[ \left(\frac{f}{g}\right)(x) = \frac{(x-6)^{2}}{1 - 3x} \] In summary: (a) \( (f+g)(x) = x^2 - 15x + 37 \) (b) \( (f-g)(x) = x^2 - 9x + 35 \) (c) \( (fg)(x) = -3x^3 + x^2 - 120x + 36 \) (d) \( \left(\frac{f}{g}\right)(x) = \frac{(x-6)^{2}}{1-3x} \)