[tex]f(x)=3x+1;\ g(x)=x^2-6\\\\(f+g)(x)=(3x+1)+(x^266)=3x+1+x^2-6=x^2+3x+(1-6)\\\\=\boxed{x^2+3x-5}[/tex]
Answer: [tex](f+g)(x)=x^2+3x-5.[/tex]
Step-by-step explanation: We are given the following two functions :
[tex]f(x)=3x+1,\\\\g(x)=x^2-6.[/tex]
We are to find the value of (f + g)(x).
We know that
for any two functions p(x) and q(x),
[tex](p+q)(x)=p(x)+q(x).[/tex]
Therefore, we get
[tex](f+g)(x)\\\\=f(x)+g(x)\\\\=(3x+1)+(x^2-6)\\\\=3x+1+x^2-6\\\\=x^2+3x-5.[/tex]
Thus, [tex](f+g)(x)=x^2+3x-5.[/tex]
