Respuesta :

In order to sketch the graph of this function, it is useful to gather the following information:

  • [tex] x=-1 [/tex] is a solution for this equation.
  • If you long divide [tex] (x^3+x^2+x+1)\div(x+1) [/tex] you get [tex] x^2+1 [/tex]. This means that we can write [tex] x^3+x^2+x+1 = (x+1)(x^2+1) [/tex]
  • Since [tex] x^2+1 [/tex] has no roots, it means that [tex] x=-1 [/tex] is the only root of this polynomial
  • This is a cubic polynomial, and the leading term has a positive coefficient. This means that [tex] \lim_{x\to\pm\infty}f(x)=\pm\infty [/tex]

So, you have to draw a cubic polynomial crossing the x axis at [tex] x=-1 [/tex], which is negative before the root and positive afterwards, and tends to negative/positive infinity as x approaches negative/positive infinity.