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

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- Jun 22nd 2006, 02:01 AMnath_quamCan 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

- Jun 22nd 2006, 03:03 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

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 - Jun 22nd 2006, 03:10 AMNanpa
What does that mean?

- Jun 22nd 2006, 03:13 AMnath_quam
Yes or No captain

- Jun 22nd 2006, 03:53 AMCaptainBlackQuote:

Originally Posted by**nath_quam**

does not mean that someone else can't.

RonL - Jun 22nd 2006, 08:38 AMThePerfectHacker
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. - Jun 23rd 2006, 05:27 AMnath_quamThanks 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

- Jun 23rd 2006, 08:57 AMSoroban
Hello, nath_quam!

I won't claim that your teacher is totally wrong,

. . but his reasoning worries me . . .

Quote:

. . . but my teacher says due to the fact at x = 1 there is multiple roots**??**

therefore there must be a point of inflection.

*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.

Quote:

or another method is to separate the equation into two graphs and*add***?**the ordinates.

**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? - Jun 24th 2006, 04:06 AMnath_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

- Jun 24th 2006, 04:51 AMSoroban
Hello, nath_quam!

Quote:

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

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.