# What are all the functions that satisfy (Calculus II)

1. Mar 4, 2008

### mnech

The problem:
$$(\int f(x)dx)(\int \frac{1}{f(x)} dx) = -1$$
solve for f(x)

The attempt at a solution

divide both sides by integral(f(x)dx):
$$\int \frac{1}{f(x)} dx = \frac{-1}{\int f(x)dx}$$

took derivative of both sides:
$$\frac{1}{f(x)} = \frac{f(x)}{(\int f(x) dx)^2}$$

multiplied both sides by f(x):
$$1 = \frac{f(x)^2}{(\int f(x) dx)^2}$$

square root:
$$1 = \frac{f(x)}{\int f(x) dx}$$

Now, here I applied L'Hopital's incorrectly (this isn't even a limit I know!) by taking derivative of top and bottom on the right side; it seemed to work later that's why I tried to stick with it:
$$1 = \frac{f'(x)}{f(x)}$$

took antiderivative of both sides...
$$x = ln|f(x)|$$

$$e^x = |f(x)|$$

$$f(x) = e^x and -e^x$$

when plugging those two answers into the original equation it seems to work...
but something isn't right and anyway there is no way that only those two functions satisfy that. help?

(first time trying to use the tex tags so i have my fingers crossed that I did not make a mistake in formatting..)

Last edited: Mar 4, 2008
2. Mar 4, 2008

### Dick

When you took an antiderivative there is a constant of integration possible. So ln(|f(x)|)=x+C. Now what? Also when you took the square root, the square root of 1 could also be -1. That gives you more solutions.

Last edited: Mar 4, 2008
3. Mar 4, 2008

### Dick

And actually the use of l'Hopital is completely incorrect. It's not even a limit. Do it this way. When you get to integral(f(x))^2=f(x)^2, just differentiate both sides and cancel the f(x). Then differentiate again. Much simpler. Now you just have the differential equation d^2(f(x))/dx^2=f(x).

4. Mar 4, 2008

### mnech

However, I am unsure how a -1 will give me more answers. when taking the square root of both sides, it would become:

$$\frac{|f(x)|}{|\int f(x) dx|} = (negative?)1$$

sorry, but can you elaborate, by including -e^x in the answer (from the absolute value another step or two down), how is that excluding anything?

As far as the constants go, we could bring them both to the right side, bringing us $$e^{ln|f(x)| + c} = e^c * |f(x)|$$ we could just call that constant $$e^c = k$$ bring it to the left side thus making the answer $$ke^x and -ke^x$$

Last edited: Mar 4, 2008
5. Mar 4, 2008

### Dick

Um, how about e^(-x)?? You are missing stuff related to those. Like I said, if you reduce the problem to f''=f there are ODE type techniques to handle that.

6. Mar 4, 2008

### mnech

ODE type techniques?

7. Mar 5, 2008

### Dick

Ordinary Differential Equations. More systematic ways of doing the same thing you've been doing.