I have come across this question whilst doing my homework and would like to clarify on the value of m.
The median of a continous random variable is given below.
$\int_{-\infty}^{m}f(x)dx = \frac{1}{2}$
Suppose $f(x) = \frac{1+\alpha x}{2}$ where $-1 \le x \le 1$ and $-1\le \alpha \le 1$, find the median of the distribution in terms of $\alpha$.
I calculated $F(x) = \int_{-1}^{x}\frac{1+\alpha x}{2} = \frac{1}{2}x + \frac{\alpha}{4} x^2 + \frac{1}{2} - \frac{\alpha}{4}$ , then $F(m) = \frac{1}{2}$
I got this formula $m = \frac{-1}{\alpha} + \sqrt{1+\frac{1}{\alpha^2}}$ or $m = \frac{-1}{\alpha} - \sqrt{1+\frac{1}{\alpha^2}}$ but I'm not sure which expression to choose.
But given that $-1\le x \le 1 $ then $-1\le m\le1$ but when $\alpha = 0$, the expression for m is invalid.
I would like to clarify then how I'm suppose to write the median value in terms of $\alpha$ .
Look at it this way. If $\alpha > 0$, then $f(x)$ is an increasing function of $x$, meaning that it will have higher density for positive values of $x$. Hence the median of $X$ will also be positive. Conversely, if $\alpha < 0$, then $f(x)$ is a decreasing function of $x$, and the median will be negative. And when $\alpha = 0$, then $f$ is uniform on $[-1,1]$ and the median will be $0$.
Now look at the expression you got for the median $m$. For all $\alpha \ne 0$, $$\left|\frac{1}{\alpha}\right| = \sqrt{\frac{1}{\alpha^2}}.$$ So $$\sqrt{1 + \frac{1}{\alpha^2}} > \sqrt{\frac{1}{\alpha^2}} = \left|\frac{1}{\alpha}\right|.$$ Consequently, $$-\frac{1}{\alpha} + \sqrt{1 + \alpha^2} > 0$$ obviously if $\alpha < 0$ (since all terms are positive), and is still positive if $\alpha > 0$ because of the inequality we just showed. Thus this first root for $m$ is always positive for any $\alpha \ne 0$. It can't be the median if $\alpha < 0$.
In a similar vein, the root $$-\frac{1}{\alpha} - \sqrt{1 + \frac{1}{\alpha^2}}$$ is always negative for any $\alpha \ne 0$. So this can't be the median if $\alpha > 0$.
We can summarize all of this nicely by writing
$$m = -\frac{1}{\alpha} + \operatorname{sign}(\alpha) \sqrt{1 + \frac{1}{\alpha^2}},$$
where $\operatorname{sign}(\alpha)$ is the sign of $\alpha$ (i.e., it is $1$ if $\alpha > 0$ and $-1$ if $\alpha < 0$.