Draw the graph of `y=x^(3)-x^(2)+x-2` and find the number of real root(s) of the equation `x^(3)-x^(2)+x-2=0`. Also locate the root.

Explanation

We have `y=f(x)=x^(3)-x^(2)+x-2`
`(dy)/(dx)=3x^(2)-2x+1 gt 0, AA x in R`
So `y=f(x)` increases.
Since the leading coefficient is '3', when `x rarr oo, f(x)rarr oo` and when `xrarr-oo, f(x)rarr-oo`.
`f(0)=-2`, hence the graph cuts the x-axis for some positive value of x.
Thus, the graph will cut the x-axis at one point only.
Also `f(1)=-1` and `f(2)=4`.
Thus, the graph will cut the x-axis somewhere between '1' and '2'.
From the above discussion, the graph of `y=f(x)` can be drawn as shown in the following figure.