Question

II Work out question 1) Let f(x) = 2x + 1 and g(x)=-3x-4: then Determine a) (f+g) (x) b) (f-g) (x) c) (f-g) (x) d) (f/g) (x) ​

Answer 1

These are all abbreviated ways of writing a sum or product of functions.

$(f+g)(x) = f(x) + g(x)$

$(f\times g)(x) = f(x) \times g(x)$

Given $f(x) = 2x+1$ and $g(x) = -3x-4$, we have

a)

$(f+g)(x) = (2x+1) + (-3x-4) = (2x-3x) + (1-4) = \boxed{-x-3}$

b)

$(f-g)(x) = (2x+1) - (-3x-4) = (2x + 3x) + (1 + 4) = \boxed{5x+5}$

c) same as (b), but I bet you meant to use some other symbol. I'll just assume multiplication:

$(f\times g)(x) = (2x+1)\times(-3x-4) \\\\ = 2x\times(-3x)+1\times(-3x) +2x\times(-4) + 1\times(-4) \\\\ = -6x^2 - 3x - 8x - 4 \\\\ = \boxed{-6x^2 - 11x - 4}$

d)

$\left(\dfrac fg\right)(x) = \dfrac{2x+1}{-3x-4} = \boxed{-\dfrac{2x+1}{3x+4}}$

though you could go on to simplify the quotient via long division; you would end up with the equivalent function (assuming x ≠ -4/3)

$\left(\dfrac fg\right)(x) = -\dfrac{2x+1}{3x+4} = -\dfrac23 + \dfrac5{3(3x+4)}$