Given that \( f(x)=x^{2}-4 x \) and \( g(x)=x+4 \), find (a) \( (f+g)(x)= \) (b) \( (f-g)(x)= \) प (c) \( (f g)(x)= \) (d) \( \left(\frac{f}{g}\right)(x)= \) Question Help:

To solve the given problems, let's start by calculating each of the required operations: Given: - \( f(x) = x^2 - 4x \) - \( g(x) = x + 4 \) (a) For \( (f+g)(x) \): \[ (f+g)(x) = f(x) + g(x) = (x^2 - 4x) + (x + 4) = x^2 - 3x + 4 \] (b) For \( (f-g)(x) \): \[ (f-g)(x) = f(x) - g(x) = (x^2 - 4x) - (x + 4) = x^2 - 5x - 4 \] (c) For \( (fg)(x) \): \[ (fg)(x) = f(x) \cdot g(x) = (x^2 - 4x)(x + 4) = x^3 + 4x^2 - 4x^2 - 16x = x^3 - 16x \] (d) For \( \left(\frac{f}{g}\right)(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2 - 4x}{x + 4} \] This can be simplified: \[ = \frac{x(x-4)}{x+4} \] It’s important to note that this expression is undefined when \( g(x) = 0 \), i.e., when \( x = -4 \). So the final answers are: (a) \( (f+g)(x) = x^2 - 3x + 4 \) (b) \( (f-g)(x) = x^2 - 5x - 4 \) (c) \( (fg)(x) = x^3 - 16x \) (d) \( \left(\frac{f}{g}\right)(x) = \frac{x(x-4)}{x+4} \) (undefined at \( x = -4 \))