# Roots of a given f(x)

1. Jul 27, 2009

### soandos

Is it possible to find a root of f(x), given just f(x_1)==y and f ' (x)?
if so, how would one go about it?
If this is in the wrong forum can a mod please move it?
thanks.

2. Jul 27, 2009

### Feldoh

Yes, by the fundamental theorem of calculus.

3. Jul 27, 2009

### soandos

4. Jul 27, 2009

### Feldoh

$$\int_{x_1}^{x} f'(t) dt = f(x)-f(x_1)$$

So effectively you have found f(x).

5. Jul 27, 2009

### soandos

so solve that for x?

6. Jul 27, 2009

### Feldoh

Why do you think you would solve for x? How would you normally find the roots of an equation once you know what the equation is?

7. Jul 27, 2009

### soandos

Im sorry, im confused.
I think to solve for x, as that was the original variable.
I generally solve equations by setting them to zero and then using inverses.
I was asking for a different method.

8. Jul 27, 2009

### Feldoh

Oh I see what your saying -- sorry I misunderstood.

You're saying solve for x such that f(x) = 0 right?

9. Jul 27, 2009

### soandos

yes.

10. Jul 28, 2009

### HallsofIvy

Staff Emeritus
Feldoh assumed you meant you were given the value of f at the single point $x_1$ and the derivative for all x. His point was that you can then integrate the derivative, using that one point to determine the constant of integration, to recover the function itself:
$$f(x)= \int_{x_1}^x f'(x)dx+ y[/itex] Set that equal to 0 and solve. Probably the best way to solve that equation, given the information you have, would be the Newton-Raphson Algorithm. Starting with any convenient value of x, perhaps $x_1$, Do [tex]x_{n+1}= x_n- \frac{f(x_n)}{f'(x_n}= x_n-\frac{\int_{x_1}^{x_n} f'(x)dx+ y}{f'(x)}$$
repeatedly until two succesive values are within your error tolerance of each other.

11. Jul 28, 2009

### soandos

But is there a way to get exact roots using that information?

12. Jul 28, 2009

### Office_Shredder

Staff Emeritus
In general you can't invert functions/find roots. For many useful specific cases, you can. Whether you can find exact roots depends on what your function f'(x) is

13. Jul 28, 2009

### soandos

here was my thinking.
a root is determined by f(x)=0
in general, f(x_1)=y, so f(x)-y = 0
can the derivative of f(x) tell us how much x had to change.
If this was simple slope, it would be easy, distance to change/slope.
cant seem to do it with the derivative.
I also know that there is a way to solve for the roots of an n degree polynomial exactly, but i don't quite get it. thing is, it shows that its always possible even when the function has no inverse.
yes i realize that broadening it to f(x) is not quite the same thing
wondering if its possible though.

14. Jul 28, 2009

### HallsofIvy

Staff Emeritus
Then you know something no mathematician knows! It was proved, by Galois in the nineteenth century, that there cannot exist a formula for solutions of polynomials of degree five or greater.

Thing is, you are completely wrong.

15. Jul 28, 2009

### soandos

Last edited: Jul 28, 2009
16. Jul 28, 2009

### Count Iblis

Yes, it is easy to derive a series expansion or write down the differential equation the inverse function has to satisfy.

17. Jul 28, 2009

### soandos

How is that done?