# Prove that this equation has at least one real root

1. Jun 11, 2013

### utkarshakash

1. The problem statement, all variables and given/known data
Let f:R→R be a continuous and differentiable function, then prove that the equation f'(x)+λf(x)=0 has at least one real root between any pair of roots of f(x)=0, λ being a real number

2. Relevant equations

3. The attempt at a solution
All that I know from Rolle's Theorem is that between a pair of roots of f(x) there must be atleast one root of f'(x). But I can't figure out how to deal with that extra term 'λf(x)'?

2. Jun 11, 2013

### dirk_mec1

Hint: f'(x) = -λf(x).

3. Jun 11, 2013

### shortydeb

Do you mean f is continuously differentiable from R to R? No textbook would say a function is continuous and differentiable on a given interval because on a given interval, a differentiable function is always continuous.

4. Jun 11, 2013

### haruspex

That is not provided as a differential equation. The question could be worded better. It is defining a function g(x) = f'(x)+λf(x), and asks you to show that g has a root between each pair of roots of f.

5. Jun 11, 2013

### haruspex

Hint 1: Does the form f'+λf remind you of anything?
Hint 2: If h(x) has no roots then f(x)h(x) has the same roots as f(x).

6. Jun 11, 2013

### Dick

You might also imagine what the graph of log(|f|) looks like on the interval and then think about how that might relate to your problem. That might give you some intuition about the problem.

7. Jun 12, 2013

### utkarshakash

Is it related to differential equations?

8. Jun 12, 2013

### haruspex

Yes.

9. Jun 12, 2013

### utkarshakash

OK then it's time for me to wait a little because my teacher hasn't started DE yet. I wonder why he gives questions which involves DE's.

10. Jun 12, 2013

### Dick

You don't have to use differential equations. Draw a sample function and sketch the graph of log(|f|).

11. Jun 12, 2013

### Saitama

Solving the given differential equation, $f(x)=ce^{-\lambda x}$ (where c is a constant). How will this function have any roots (except $\infty$)?

12. Jun 12, 2013

### verty

This fooled me too. If g(x) = 0 for all x, THEN f is that function (is in that family).

13. Jun 12, 2013

### verty

Hint 3: will it suffice to consider only pairs of adjacent roots of f?

14. Jun 12, 2013

### haruspex

As I pointed out to dirk_mec1, they have not provided us with a differential equation for f.
Let me reword the question to make it clearer:
(It's wrong to talk about an equation having roots. Functions have roots, equations have solutions. A root of f(x) is a solution of f(x)=0.)
That said, you have indeed found a function that has no roots, and it is the function h(x) in my second hint.

15. Jun 12, 2013

### shortydeb

I don't know if the proposition given in the OP is true if f is differentiable but not continuously differentiable, but you can use the Intermediate Value Theorem to prove the proposition if f is continuously differentiable.

Last edited: Jun 12, 2013
16. Jun 13, 2013

### verty

I was just noticing that a function like $f(x) = e^{-\frac{1}{x^2}} * e^{-\frac{1}{(x-1)^2}}, f(0) = f(1) = 0$ is a problem for this type of argument.

17. Jun 13, 2013

### CompuChip

Is that differentiable at x = 0, x = 1?

18. Jun 13, 2013

### epenguin

Either I am missing something or (as it seems to me) a big meal is being made of something damned obvious. The question does assume that f have two real roots, otherwise 'between' makes no sense.

So just consider what f'(x)+λf(x) is at one root of f and the next root of f.

19. Jun 13, 2013

### shortydeb

Look at any pair of roots. Consider the case where f'(x)+λf(x) is positive at the first root and negative at the second root. What does the Intermediate Value Theorem tell you about f'(x)+λf(x) between the two roots? This is assuming f is continuously differentiable on R.

20. Jun 13, 2013

### utkarshakash

Since f'(x)+λf(x) has changed its sign from +ve to -ve there must be atleast one point where it became zero. Is this logic correct?