# Finding a solution

1. Nov 23, 2008

### Firepanda

Show that the equation xe^x = 1 has a solution between 0<x<1.

Well for this I simply changed to x = 1/e^x

Iterated starting with x=1, and got 0.567

Then it asks, prove that the solution is unique, which I guess means there is no other solution for the equation but 0.567.

2. Nov 23, 2008

### Firepanda

Ok perhaps wrong, another way:

There is a solution where x0 is an element of [0,1]

where f(x) = xe^x - 1

f(0) = -1 < 0 < e^1 - 1 = f(1)

By the intermediate value theroem there exists an x0 in the interval with f(x0) = 0.
Correct?

Still unsure on the second part, think it uses Rolle's theorem.

3. Nov 23, 2008

### Firepanda

Think I have it:

For a,b in the interval, where a is not equal to b, then f(a) can not be equal to f(b) if x0 is unique.

Then suppose f(a) does equal f(b), apply Rolles theorem:

=> there is an x0 in the interval where f'(x0) = 0

But f'(x) = xe^x + e^x which can never be equal to zero since it is always positive for our interval [0,1] and all positive values of x, which is a contradiction.

Now is the question asking me if the result is unique in the interval or for all x?

Edit: So f'(x) is always posive for [-1,inf). Am i supposed to show the lim x-> -inf or something?

Last edited: Nov 23, 2008
4. Nov 24, 2008

### HallsofIvy

Staff Emeritus
yes, what you have done is perfectly good.

5. Nov 24, 2008

### Firepanda

Is there a way to show that the solution is unique for all x? i can graph it and see it is, it's monotonic for [-1,inf), do i have to show for all (-inf,-1) that f(x) is always negative?

6. Nov 24, 2008

### Staff: Mentor

Since you have shown that f(0) = -1, f(1) > 0, and f'(x) > for x in [0, 1], so f(x) = 0 for only a single value. A function whose derivative is positive on an interval can't cross the x-axis more than once on that interval.