# I don't know how to calculate it: x^3-2x^2+x-1

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### 2 Answers

You should use the derivative of the function to evaluate the intervals that contain the real roots of the equation `x^3-2x^2+x-1 = 0` such that:

`f'(x) = (x^3-2x^2+x-1)' => f'(x) = 3x^2 - 4x + 1`

You need to solve for x the equation `3x^2 - 4x + 1 = 0` such that:

`3x^2 - 4x + 1 = 0`

`x_(1,2) = (4+-sqrt(16 - 12))/6 => x_(1,2) = (4+-2)/6`

`x_1 = 1 ; x_2 = 1/3`

You should evaluate the value of the function at the critical values `x=1` and `x=1/3` such that:

`f(1) = 1^3-2+1-1 = -1`

`f(1/3) = 1/27 - 2/9 + 1/3 - 1=> f(1/3) = (1-6+9-27)/27`

`f(1/3) = -23/27`

**Notice that the sign does not changes over intervals `(-oo,1/3), (1/3,1)` but it changes over `(1,oo)` , hence, the function has one real root over `(1,2)` and it does have two complex conjugate roots.**

This is a cubic equation because it contains x^3 in it. The graph is an odd function because it has symmetry when you rotate it by 180 degrees about the origin. Cubic equations have at most three routes. Put in numbers for x to get output values for y. Plot on a graph to see what the function looks like.

To look for integer roots look at factors of the constant term on the end, -1.

Integer factors of -1 are 1 and -1.

Put 1 into the equation to try it. We get 1^3 - 2*1^2 + 1 - 1 = 1 - 2 + 1 - 1 = -1. Not equal to zero, so not a root!

Put -1 into the equation. We get (-1)^3 - 2*(-1)^2 + 1 -1 = -1 -2 +1 -1 = -3. Oh dear, not a root either!

If you draw the function you will see that it cuts the x axis only once so there is only one real factor. The other two factors are complex.

To draw the graph without a calculator, work out where it cuts the y-axis. This is the constant term on the end, -1. To find the turning points, differentiate and set to zero. There are turning points at x=1/3 and x=1. Put these values in to get the values of y.