# Finding y=f(x)

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1. Apr 29, 2015

1. The problem statement, all variables and given/known data
Given two curves y=f(x) passing through (0,1) and $g(x)=\int\limits_{-\infty}^xf(t)dt$ passing through (0,1/n). The tangents drawn to both curves at the points with equal abscissae intersect on the x-axis. Find y=f(x).

2. Relevant equations
None

3. The attempt at a solution
g(0)=$\int\limits_{-\infty}^0f(x)dx$=1/n
let the abscissae be x.
The tangent to y=f(x) is y=xf'(x)+c
Can I directly differentiate g(x) to get slope?

2. Apr 29, 2015

### HallsofIvy

Staff Emeritus
By the "fundamental theorem of Calculus", $g'(x)= f(x)$

3. Apr 29, 2015

What about the lower limit? So differentiating $\int\limits_a^xf(t)dt$ with respect to x gives the same value for any a?

4. Apr 29, 2015

Ok. Let integral of f(t) from a to x be F(x)-F(a). Since F(a) is a constant, g'(x) is f(x). Thank you for helping.

5. Apr 29, 2015

### HallsofIvy

Staff Emeritus
Yes. For any a and b, $\int_a^x f(x)dx= \int_b^x f(x)dx+ \int_a^b f(x)dx$ and $\int_a^b f(x)dx$ is a constant.