Answer:
[tex](f+g)(x) = 3x^2+\frac{3x}{2}-9\\or\\(f+g)(x) = \frac{6x^2+3x-18}{2}[/tex]
Step-by-step explanation:
We are given:
[tex]f(x)=\frac{x}{2}-3 \,\, and\,\, g(x) = 3x^2+x-6[/tex]
We need to find [tex](f+g)(x)[/tex]
(f+g)(x) can be found by adding f(x) and g(x)
(f+g)(x) = f(x) + g(x)
[tex](f+g)(x) = \frac{x}{2}-3+(3x^2+x-6) \\(f+g)(x) = \frac{x}{2}-3+3x^2+x-6\\(f+g)(x) = 3x^2+\frac{x}{2}+x-3-6\\(f+g)(x) = 3x^2+\frac{3x}{2}-9\\(f+g)(x) = \frac{6x^2+3x-18}{2}[/tex]
so, (f+g)(x) is:
[tex](f+g)(x) = 3x^2+\frac{3x}{2}-9\\or\\(f+g)(x) = \frac{6x^2+3x-18}{2}[/tex]
