Respuesta :

konrad509
[tex]\int 6x+18\, dx=3x^2+18x+C\\\\ 3=3\cdot 1^2+18\cdot1+C\\ 3=3+18+C\\ C=-18\\ f'(x)=3x^2+18x-18\\\\ \int 3x^2+18x-18\, dx=x^3+9x^2-18x+C\\\\ 5=1^3+9\cdot1^2-18\cdot1+C\\ 5=1+9-18+C\\ C=13\\\\ f(x)=x^3+9x^2-18x+13 [/tex]

[tex]f(-2)=(-2)^3+9\cdot(-2)^2-18\cdot(-2)+13\\ f(-2)=-8+36+36+13\\ \boxed{f(-2)=77}[/tex]

But honestly I don't know if it's good method.

Otras preguntas