Question 755154
More sense made by looking at {{{g(x)=x^3-x}}} because this is 1 unit less than f(x).  g(x) here is factorable and easily analyzed for its intervals regarding signs of the values for graphing of g(x).  You could then look at a graphical representation and visually judge WHERE when the graph is raised 1 unit, can exactly 1 point intersect the x axis.  


If raising the graph until the point between 0 and 1 is reached, then there still is one other point way on the left which also intersects the x axis.  Keep raising the graph until f(x) is reached, one unit above g(x).  The intersecting point is way over on the left. This is at {{{x<-1}}}.


{{{f(x)=g(x)+1}}} and {{{g(x)=x^3-x}}}
and find that {{{g(x)=x(x-1)(x+1)}}}


The intervals around the roots for g(x) are
(-infin,-1)
(-1, 0)
(0,+1)
(+1, +infin)
Check the signs of those intervals and show this on a graph, even if only qualitatively between the within each interval.
Also notice qualitatively that g(x) and therefore f(x) decrease as x tends toward lesser than -1 while g(x) has a root at x=-1.


Reminding again that the one real zero is at a particular point for {{{x<-1}}}, probably your choice (A) for -2 to -1.

f(x)
{{{graph(300,300,-3,3,-10,10,x^3-x+1)}}}


g(x)
{{{graph(300,300,-3,3,-10,10,x^3-x)}}}