Answer:
[tex]\large\boxed{(f+g)(x)=2x^2+\dfrac{1}{2}x+4\dfrac{1}{3}}[/tex]
Step-by-step explanation:
[tex](f+g)(x)=f(x)+g(x)\\\\\text{We have}\\\\f(x)=\dfrac{1}{3}-\dfrac{1}{2}x,\ g(x)=2x^2+x+4.\\\\\text{Substitute:}\\\\(f+g)(x)=\left(\dfrac{1}{3}-\dfrac{1}{2}x\right)+(2x^2+x+4)\\\\=\dfrac{1}{3}-\dfrac{1}{2}x+2x^2+x+4\qquad\text{combine like terms}\\\\=2x^2+\left(-\dfrac{1}{2}x+x\right)+\left(\dfrac{1}{3}+4\right)\\\\=2x^2+\dfrac{1}{2}x+4\dfrac{1}{3}[/tex]
