There is just one real number:
[tex]f(x) = x^3-x+1=0[/tex]
[tex]f'(x)=3x^2-1=0[/tex]
[tex]x_1=\frac{1}{sqrt(3)} x_2=-\frac{1}{sqrt(3)}[/tex]
[tex]f(\frac{1}{sqrt(3)}) = 0.615100[/tex] - minimum
[tex]f(-\frac{1}{sqrt(3)}) = 1.384900[/tex] - maximum
[tex]f(-infinity) = -infinity[/tex] it means you have a solution for f(x)=0
such solution is close to
x = -1.3247179574971
with approximation error = .0000000010761969093664
-V-
[tex]f(x)=x^3-x+1[/tex]
[tex]f(0)=1,f(-2)=-5 \Rightarrow f(0)f(-2) < 0[/tex]
By IVT, this shows that f(x) has at least one root in the interval (-2,0)
To find that root, you want to show me why Vadim's number is not right, Calvin ?