# Differentiation

• Oct 27th 2009, 01:12 AM
thekiterunner
Differentiation
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 AM
Defunkt
Quote:

Originally Posted by thekiterunner
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 $3x^2-6x+3 \geq 0$ by using the fact that $3x^2-6x+3 = 3(x^2-2x+1)$.

After you proved that, you know $f'(x) \geq 0 \ \forall x \in \mathbb{R}$. What does this tell you about the $f(x)$? What, in general, does $f'(x)$ mean in regards to $f(x)$?
• Oct 27th 2009, 01:20 PM
thekiterunner
i do not undertsand that one little bit...any chance for simplier terms
• Oct 27th 2009, 02:50 PM
hjortur
If you have a general quadratic equation:
$f(x)=ax^2+bx+c$

And let the discriminant $D=b^2-4\cdot ac$.

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 $f<0$ or $f>0$ 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 $f\leq0$ or $f\geq0$ 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 $f'(x)$ means in regard to $f(x)$.

Does that help?
• Oct 27th 2009, 03:44 PM
thekiterunner
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 PM
hjortur
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 PM
Defunkt
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 $\forall x \in \mathbb{R}, \ f'(x) \geq 0$, what does this tell you about the graph of f?