1. ## Can you determine??

On the graph p(x) = ((1-x)^3)(x-3) without the aid of calculus, graphing calculators.etc can you justify there is a point of inflection at x = 1 ??? Answer wanted arguement with maths teacher thanks

2. Originally Posted by nath_quam
On the graph p(x) = ((1-x)^3)(x-3) without the aid of calculus, graphing calculators.etc can you justify there is a point of inflection at x = 1 ??? Answer wanted arguement with maths teacher thanks
Non-rigorous argument:

Consider the related function:

pp(x)=(1-eps-x)x(1+eps-x)(x-3).

for any small eps>0. This has three roots all close to x=1, and in some
appropriate sense the average slope between x=1-eps and x=1+eps
is zero, so as eps goes to zero we generate a family of curves all of which
have an "average" slope of 0 near x=1, so it is reasonable to expect that
the slope of p(x) is 0 at x=1.

(I think I would need Roll's theorem (and the nested interval theorem?) to
make this rigorous, and your teacher would almost certainly forbid THAT).

RonL

3. What does that mean?

4. Yes or No captain

5. Originally Posted by nath_quam
Yes or No captain
If you have to ask the answer is obviously no - I can't but that
does not mean that someone else can't.

RonL

6. Here is the problem.
The definition I seen for "infecltion" is when the derivative changes from increasing to decreasing. The very definition is already based on calculus.

That means you must be using some other definition for inflection point.

7. ## Thanks but no luck

i was unable to gain full marks for the question cos i didnt include the inflexion on my sketch which we were asked to do without calculus but my teacher says due to the fact at x =1 there is multiple roots therefore there must be a point of inflexion or another method is to seperate the equation into two graphs and add the ordinates Thanks for your help

8. Hello, nath_quam!

I won't claim that your teacher is totally wrong,
. . but his reasoning worries me . . .

. . . but my teacher says due to the fact at x = 1 there is multiple roots ??
therefore there must be a point of inflection.
This is not true . . . well, okay, if the multiplicity is odd.

$\displaystyle f(x)\:=\1 - x)^2(3 - x)$ is tangent to the x-axis at $\displaystyle x = 1$

. . and has a relative maximum there.

or another method is to separate the equation into two graphs and add ? the ordinates.
The function is a product . . . what two graphs are we supposed to add?

Even if we had a rough sketch of the graph,
. . how does that prove the existence of an inflection pont?

9. By spliting the graph into two seperate graphs and sketching them and adding there ordinates it proves it or something cos one graph has a cubic curve

10. Hello, nath_quam!

By spliting the graph into two seperate graphs and sketching them and adding there ordinates
it proves it or something, cos one graph has a cubic curve
I'll try again . . . *sigh*

We have a product . . . How do we "split" it?

The function is: .$\displaystyle p(x)\;=\;(1 - x)^3(x - 3)$ . . . a product.

If you graph $\displaystyle f(x) = (1 - x)^3$ and $\displaystyle g(x) = x - 3$, . then you must multiply the ordinates.

Got it?

And I still say: Looking at a graph (even an accurate one) is not a proof.