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.
Simply solve the inequality $\displaystyle 3x^2-6x+3 \geq 0 $ by using the fact that $\displaystyle 3x^2-6x+3 = 3(x^2-2x+1)$.
After you proved that, you know $\displaystyle f'(x) \geq 0 \ \forall x \in \mathbb{R}$. What does this tell you about the $\displaystyle f(x)$? What, in general, does $\displaystyle f'(x)$ mean in regards to $\displaystyle f(x)$?
If you have a general quadratic equation:
$\displaystyle f(x)=ax^2+bx+c$
And let the discriminant $\displaystyle D=b^2-4\cdot ac$.
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 $\displaystyle f<0$ or $\displaystyle f>0$ 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 $\displaystyle f\leq0$ or $\displaystyle f\geq0$ 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 $\displaystyle f'(x)$ means in regard to $\displaystyle f(x)$.
Does that help?