f(x)=x^3-3x^2+3x
f '(x)=3x^2-6x+3
I have no issue with finding f ' (x).
The next part of the question says, thus show that f' (x) => 0 for all x and interpret this in terms of the graph f(x)=x^3-3x^2+3x
If anyone can help that would be great.
f(x)=x^3-3x^2+3x
f '(x)=3x^2-6x+3
I have no issue with finding f ' (x).
The next part of the question says, thus show that f' (x) => 0 for all x and interpret this in terms of the graph f(x)=x^3-3x^2+3x
If anyone can help that would be great.
If you have a general quadratic equation:
And let the discriminant .
Now if D<0 that means the equation has no real solution, that is the graph nowhere goes
through the x-axis. It is entirely contained on either side. Either or for all x.
Now if you dont know if f is below or above the x-axis,
just check for some point if it is above or below, then you know that the rest of the points
are on the same side.
If D=0 then there is one real solution, the graph is entirely on one side and "touches" the
x-axis in only one point, it never crosses the axis. So either or for all x.
What if D>0 ?
But anyway what is D in your equation?
After you have found out the first part, just think about, like Defunkt said, what means in regard to .
Does that help?