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

d)

`\left(\frac fg\right)(x) = (2x+1)/(-3x-4) = \boxed{-(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(\frac fg\right)(x) = -(2x+1)/(3x+4) = -\frac23 + \frac5{3(3x+4)}`