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