Inequality

$Problem$ #3 Find all real numbers $x$ for which $\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}$ Let $f(x)$ $=\sqrt{3-x}-\sqrt{x+1}$. First note that $f(x)$ is defined for $-1 \leq x \leq 3$ $f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{2\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{x+1}}\big) < 0 \Rightarrow f(x)$ is strictly decreasing Now $f(-1) = 2 > \frac{1}{2}$ and $f(3) = -2 < \frac{1}{2}$ Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$ $f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8}$ but $x = 1 + \frac{\sqrt{31}}{8}$ does not satisfy $\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}$ Check yourself! So the only solution is $x = 1 - \frac{\sqrt{31}}{8}$ Since $f(x)$ is strictly decreasing, the given inequality is true for $x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}$