ANSWER
The x-coordinate of the x-intercept of the function [tex]f(x)=x^3-x^2-x+1[/tex] is [tex]-1[/tex].
EXPLANATION:
The x-intercept refers to the point where the function meets or cuts the x-axis.
At the x-intercept, [tex]f(x)=0[/tex].
This means we have to equate the whole function to zero and solve for [tex]x[/tex].
[tex]x^3-x^2-x+1=0[/tex]
According to the rational roots theorem, the possible rational roots of the equation [tex]x^3-x^2-x+1=0[/tex] are [tex]\pm 1[/tex].
We now plug in these possible rational roots to see which of them are real roots.
[tex]f(1)=1^3-1^2-1+1[/tex]
[tex]f(1)=1-1-1+1[/tex]
[tex]f(1)=-1+1[/tex]
[tex]f(1)=0[/tex]
Since [tex]f(1)=0[/tex], [tex]x=-1[/tex] is a root.
By the factor theorem, [tex]x+1[/tex] is a factor of the function.
We now divide the polynomial function, [tex]x^3-x^2-x+1=0[/tex] by [tex]x+1[/tex] to find the remaining roots.
See long division in diagram.
This means that [tex]f(x)=(x+1)(x^2-2x+1)[/tex].
Or
[tex]f(x)=(x+1)(x-1)^2[/tex]
[tex]\Rightarrow (x+1)(x-1)^2=0[/tex]
Hence the roots are [tex]x=1[/tex] or [tex]x=-1[/tex]
In the second quadrant the x-coordinate is negative
Therefore, the x-coordinate of the x-intercept of the function [tex]f(x)=x^3-x^2-x+1[/tex] is [tex]-1[/tex].
See graph in attachment
