The curves are $y=mx$ and $y=\dfrac{x}{x^2+1}$.
I didn't know how to approach this problem, so I decided to try some numbers for $m$ in a graphing calculator. For me it looks like $m$ must be on the interval $(0,1)$. Is this correct, and is there a proof or a more mathematical argument for why this is true?
I am also asked to find the region in terms of $m$. Here's what I did: $$a=\int_{-1}^0 \left(mx-\frac{x}{x^2+1}\right)dx+\int_0^1 \left(\frac{x}{x^2+1}-mx\right)dx=\ln2-m.$$ Is that correct?
On
We want to find some point $x>0$ on the real number line such that
$$
mx = \frac{x}{x^2+1}.
$$
Note that $m>0$ so that the line and the curve will intersect at another point other than at $(0,0)$.
By clearing the denominators, the above equality implies $m(x^3+x)=x$, or
$$
x(mx^2+m-1)=0
$$
by moving all terms to one side.
So
$$
x=0 \hspace{4mm}
\mbox{ or }
\hspace{4mm}
x = \frac{-0\pm \sqrt{0^2-4m(m-1)}}{2m}
$$
by using the quadratic formula.
In order for
$$
x = \frac{-0\pm \sqrt{0^2-4m(m-1)}}{2m}
$$
to be a real number, we need what is under the radical $-4m^2+4m=4m(1-m)$ to be greater than $0$.
Since we want $4m(1-m)>0$ and $m>0$ by assumption, $1-m>0$. So $m<1$.
In conclusion, $0<m<1$.
$m$ has to be smaller than the gradient of $\displaystyle y=\frac{x}{x^2 + 1}$ at $x=0$ to enclose a region. $m$ also has to be greater than $0$ to enclose a region, otherwise it will never intersect the second graph. $\displaystyle 0<m<\frac{d}{dx}\left(\frac{x}{x^2+1}\right)$ at $x=0$, which turns out to be $1$. So, $0<m<1$.