# Could There Be A Video For This Example Practice Problem For Math

Was wondering if they could do a video on finding the function "f", that makes the above true. I feel this problem, at least to me, is good practice at pattern recognition and integration.

*** Actually I am not sure if there is an f(x) to equation above, but I think if you replace the definite integrals with indefinite integrals, perhaps then my solution is valid. ***

I figured out the solution myself, unless I made mistakes.

**C < 0, W is the Lambert W function, and e is Euler's number.**

This is my solution:

g(x) is the inverse of f(x) which is:

Here is my work:

1) Take the natural logarithm of both sides. Also since "f" is in a logarithm function, then f(?) > 0 for "a" <= ? <= x. Also, the integration of "f" from "a" to x must be greater than 0 because it equals e raised to a definite integral (assuming f(x) is a real function) which must be a positive value.

2) Take the derivative of both sides with respect to x.

3) Divide both sides by -f(x). Since f(x) > 0 < integration of "f" from "a" to x, then ln(f(x)) must be greater than 0, which means f(x) > 1.

4) Rewrite the left side of the equation.

5) Take the Lambert W function of both sides.

6) Add ln(f(x)) to both sides of the equation and then take e to the power of both sides of the equation.

7) Rewrite the right side of the equation.

8) Make a substitution to make life easier.

9) Rewrite the right side of the equation.

10) Do integration.

11) Do integration by parts for the right side.

12) Rewrite F(x).

13) Rewrite the right side of the equation.

14) Rewrite the right side of the equation.

15) Rewrite the right side of the equation.

16) Rewrite the right side of the equation. Because W(-1/F) < 0, then |W(-1/F)| = -W(-1/F)

17) Rewrite both sides of the equation.

From here it is just algebra. Since I used up the image memory limit, I will try my best to write the rest down.

18) Make a substitution to make life easier.

z = ln(-W(-1/F))

19) Rewrite the right side of the equation.

x + C = z + e ^ -z

20) Subtract z from both sides of the equation.

(x + C) - z = e ^ -z

21) Multiply both sides of the equation by -e ^ (z - x - C).

(z - x - C)) * e ^ (z - x - C)) = -e ^ (-x - C)

22) Take the Lambert W function of both sides of the equation.

z - x - C = W(-e^(-x - C))

23) Add x + C to both sides of the equation.

z = x + C + W(-e^(-x - C))

24) Rewrite z.

ln(-W(-1/F)) = x + C + W(-e^(-x - C))

25) Take e raised to the power of both sides of the equation.

-W(-1/F) = e ^ C * e ^ x * e ^ W(-e ^ (-x - C))

26) Rewrite the right side of the equation.

-W(-1/F) = -1 / W(Ce^-x), **C < 0 ***Assume this is true for the following. Also because W(t) is undefined in the Reals when t < -1/e, then F(x) >= e.*

27) Remember that W(-1/F) = -ln(f(x)), so do substitution for W(-1/F).

-1 / W(Ce^-x) = -W(-1/F) = ln(f(x))

-1 / W(Ce^-x) = ln(f(x))

28) Raise e to power of both sides of the equation.

f(x) = e ^ (-1 / W(Ce^-x))

Please sign in to leave a comment.