Respuesta :

caylus
Hello,

[tex]f(x)=2x^2+2x+3\\ f(3)=2*3^2+2*3+3=18+6+3\\ f(3+h)=2(3+h)^2+2*(3+h)+3\\\\ =2(9+6h+h^2)+6+2h+3\\\\ =2h^2+14h+18+6+3\\\\ \dfrac{f(3+h)-f(3)}{h} = \dfrac{2h^2+14h+18+6+3-(18+6+3)}{h} \\ = \dfrac{2h^2+14h}{h} )=2h+14 [/tex]