Finding y=f(x) with Tangents and Equal Abscissae Intersection

[AdityaDev]

Homework Statement

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).

Homework Equations

None

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?

 

[HallsofIvy]

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

 

[AdityaDev]

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

 

[AdityaDev]

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.

 

[HallsofIvy]

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

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.