- #1

- 20

- 0

A = set of all x such that f(x) = 0

B = set of all x such that f'(x) = 0

then is there any pattern to finding A if B is known (or vice versa)?

Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter drewfstr314
- Start date

- #1

- 20

- 0

A = set of all x such that f(x) = 0

B = set of all x such that f'(x) = 0

then is there any pattern to finding A if B is known (or vice versa)?

Thanks!

- #2

Bacle2

Science Advisor

- 1,089

- 10

Mostly so for functions of the form f(x)=(x-a)^n , but there is no necessary

dependence for all functions: look at, e.g., f(x)=sinx , which is 0 at k*Pi ,

but f'(x)=cosx is 0 at (2n+1)*Pi/2 . If f(x)=f'(x)=0 , then f has both a 0 and

a critical point at x . Then you can shift your function by a constant and the

relationship does not exist any more-- the zero is no more, has been shifted.

- #3

chiro

Science Advisor

- 4,797

- 133

IMO there are two important classes to think about: the first is where you have a true polynomial and the second is when you don't.

The examples of non-polynomials include anything with a proper power series where you have all non-zero co-effecients (or even some kind of shifted power series like a taylor series expansion with a non-zero center).

In the finite-case you should use the relationship for the anti-derivative of a*x^n and use the fact that the original equation can be written in terms of (x - c)(x - d)...(x - last_root) = 0, after which you relate the original expression to a new expression (i.e the antiderivate) in any way you can.

In the infinite-case you can't quite do the same thing, but a suggestion for this is to look at the exponential function and use a composition of functions and see how the deriatives change.

For the finite case, you might want to consider trying the following: suppose you have a integral expression for the roots of the integral (in other words, (x - a)(x - b)...(x - n) = 0 for the integral not the derivative). Now you can use the product rule to differentiate this expression and get something in terms of the derivative and how it should equal to 0.

You also have the reverse situation where you can start from the roots of the derivative (i.e (x - a)(x - b)...(x - n) = 0 but you have one less x and the coeffecients are different).

So basically the point I'm making is that you can generate quite a few identities for both equations and then equate the right ones together to get expression in terms of the other.

It's not obviously an answer, but the idea of mathematics is to use as many independent kinds of data as you can and then to bring them all together: the more linearly independent pieces of information you have, the more choices you have to analyze the said problem and the more you can do with it.

Having at least two inter-changeable non linearly independent pieces is what you need and in some cases, this is referred to as a duality. If you have dualities that have their own dualities then you get more and more independent expressions of the same thing and this generates a systematic way to analyze something since you are generating lots of independent ways of describing something.

- #4

- 56

- 0

- #5

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 967

- #6

- 20

- 0

[itex]f(x) = x^3 - 4x + 2 \Rightarrow f'(x) = 3x^2 - 4[/itex]

and the solutions of f'(x)=0 are [itex]x = \pm \frac{2\sqrt3}{3}[/itex]

Based on this, is there any way to find the zeros of f(x)?

Thanks!

- #7

- 22,129

- 3,298

[itex]f(x) = x^3 - 4x + 2 \Rightarrow f'(x) = 3x^2 - 4[/itex]

and the solutions of f'(x)=0 are [itex]x = \pm \frac{2\sqrt3}{3}[/itex]

Based on this, is there any way to find the zeros of f(x)?

Thanks!

I don't think you can use this to find the zeros of f(x). The graph shows that f has 3 zeros. So all you can say (in this case), is that you got three zeros a, b and c and these must satisfy [itex]a\leq - \frac{2\sqrt{3}}{3}\leq b\leq \frac{2\sqrt{3}}{3}\leq c[/itex].

I don't think you can do much more.

- #8

pwsnafu

Science Advisor

- 1,080

- 85

[itex]f(x) = x^3 - 4x + 2 \Rightarrow f'(x) = 3x^2 - 4[/itex]

and the solutions of f'(x)=0 are [itex]x = \pm \frac{2\sqrt3}{3}[/itex]

Based on this, is there any way to find the zeros of f(x)?

Thanks!

Short answer no. For the simple reason that ##g(x) = x^3 -4x + 102## has the same derivative as ##f##, but they don't have the same roots.

- #9

- 10

- 0

If we say that gravitatory field is proportional to the square of the distance for the tridimensionality of space, I think law of our gravitatoryish is explained by the bidimensionality of the complex plane.

Share: