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.

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- Oct 27th 2009, 01:12 AMthekiterunnerDifferentiation
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. - Oct 27th 2009, 01:50 AMDefunkt
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 $\displaystyle f(x)$? - Oct 27th 2009, 01:20 PMthekiterunner
i do not undertsand that one little bit...any chance for simplier terms

- Oct 27th 2009, 02:50 PMhjortur
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? - Oct 27th 2009, 03:44 PMthekiterunner
But anyway what is D in your equation?

D=6^2 - (4 x 3 x 3) = 0

So becuase D=0 i have one real solution

f'(x) is just the gradient function of f(x) - Oct 27th 2009, 04:42 PMhjortur
So if f'(x) is always positive or zero, then f(x) must be increasing (not strictly).

And f(x) does not have a local maxima or minima at it's critical point. - Oct 27th 2009, 05:12 PMDefunkt
The geometrical meaning of f'(x) regarding f(x) is the slope -- f'(x0) is (if it exists) the slope of the function f in point x0.

Now, if $\displaystyle \forall x \in \mathbb{R}, \ f'(x) \geq 0 $, what does this tell you about the graph of f?