Homework Problem Statement:
Any line of the form $y = mx$ will intersect the curve $y = \frac {x}{x^2+10} $ in precisely three points provided $0 < m < B$ for some number $B$. What is the value of $B$?
When the line $y = mx$ intersects the curve in three points, this line and the curve $y = \frac {x}{x^2+10} $ will enclose a region consisting of two parts. Find the total area of the enclosed region. Your answer should be an expression involving $m$.
For the first part, I set the above equations to equal each other to find what value of $B$ will be greater than $m$. $B$ equals $\frac {1}{10}$, therefore $0 < m < \frac {1}{10}$.
As for the second part, I am unsure of what to do. My first approach is to apply the same method from the first part, but this time, to solve for $x$. I have found that when $mx = \frac {x}{x^2+10}$, $x = \sqrt\frac {1-10m}{m}$ and $x = -\sqrt\frac {1-10m}{m}$.
The area under the curve is $A = \int f(x) - g(x)\ \textrm dx$.
From $-\sqrt\frac {1-10m}{m}$ to $0$, $f(x) = mx$ and $g(x) = \frac {x}{x^2+10} $.
From $0$ to $\sqrt\frac {1-10m}{m}$, $f(x) = \frac {x}{x^2+10}$ and $g(x) = mx$.
After finding the area between those two, I added them together to find the total area. However, my answer was not correct.
edit: I apparently misplaced a parenthesis when inputting my answer. The answer is 10m + ln ($\frac {1}{10m$}$)-1
