Respuesta :

Let [tex]f(x)=x^5-(x+1)[/tex]. Then [tex]f(1)=-1[/tex] and [tex]f(2)=29[/tex]. By the intermediate value theorem, it follows that there is some [tex]c\in(1,2)[/tex] such that [tex]f(c)\in[f(-1),f(2)]=[-1,29][/tex].

This guarantees that there is some [tex]c[/tex] between -1 and 2 such that [tex]f(c)=0[/tex], i.e. there is some [tex]c[/tex] such that [tex]c^5=c+1[/tex].