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: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
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
Hello, nath_quam!
I won't claim that your teacher is totally wrong,
. . but his reasoning worries me . . .
This is not true . . . well, okay, if the multiplicity is odd.. . . but my teacher says due to the fact at x = 1 there is multiple roots ??
therefore there must be a point of inflection.
1 - x)^2(3 - x)" alt="f(x)\:=\1 - x)^2(3 - x)" /> is tangent to the x-axis at
. . and has a relative maximum there.
The function is a product . . . what two graphs are we supposed to add?or another method is to separate the equation into two graphs and add ? the ordinates.
Even if we had a rough sketch of the graph,
. . how does that prove the existence of an inflection pont?
Hello, nath_quam!
I'll try again . . . *sigh*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: . . . . a product.
If you graph and , . then you must multiply the ordinates.
Got it?
And I still say: Looking at a graph (even an accurate one) is not a proof.