Question

resolva
f(x)= x³+2x+3
g(h)= [f(3+h) - f(3)] ÷ h
g(h)= ?

Answer 1

$g(h)~=~\frac{f(3+h)-f(3)}{h}\\\\\\g(h)~=~\frac{\left((3+h)^3~+~2\,.\,(3+h)~+~3\right)~-~f\left(3\right)}{h}\\\\\\g(h)~=~\frac{\left((h^3+9h^2+27h+27)~+~(6+2h)~+~3\right)~-~f\left(3\right)}{h}\\\\\\g(h)~=~\frac{\left(h^3+9h^2+27h+27~+~6+2h~+~3\right)~-~(3^3~+~2\,.\,3~+~3)}{h}\\\\\\g(h)~=~\frac{\left(h^3+9h^2+27h+27~+~6+2h~+~3\right)~-~(27~+~6~+~3)}{h}\\\\\\g(h)~=~\frac{h^3+9h^2+27h+27~+~6+2h~+~3~-~27~-~6~-~3}{h}\\\\\\g(h)~=~\frac{h^3+9h^2+29h}{h}$

$g(h)~=~\frac{h\,.\,(h^2~+9h+29)}{h}\\\\\\g(h)~=~\frac{1\,.\,(h^2~+9h+29)}{1}\\\\\\\boxed{g(h)~=~h^2+9h+29}$