Restricting a sinusoid so it has an inverse

A function has an inverse on an interval when it takes each output value only once there. For a sinusoid, the natural choice is a stretch from a minimum to the next maximum, where it rises steadily through every value it can reach.

Find those points. $g(x)=\sin x-\cos x$ can be rewritten as $\sqrt{2}\sin (x-\frac{\pi }{4})$, so its range is $-\sqrt{2}\le g(x)\le \sqrt{2}$. It is smallest when $x-\frac{\pi }{4}=-\frac{\pi }{2}$, at $x=-\frac{\pi }{4}$, and largest when $x-\frac{\pi }{4}=\frac{\pi }{2}$, at $x=\frac{3\pi }{4}$. Check: $g(-\frac{\pi }{4})=-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$ and $g\left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$.

Between these, $g$ increases the whole way and hits every value from $-\sqrt{2}$ to $\sqrt{2}$ once. That is choice B, with its stated reason.

Each choice

• A. On $0\le x\le \pi$, $g$ starts at $-1$, climbs to $\sqrt{2}$ at $\frac{3\pi }{4}$, then falls back to 1. Values between 1 and $\sqrt{2}$ appear twice and values below $-1$ never appear, so the stated reason is false here.
• B (correct). From the minimum at $-\frac{\pi }{4}$ to the maximum at $\frac{3\pi }{4}$, $g$ rises through every value from $-\sqrt{2}$ to $\sqrt{2}$ exactly once.
• C. Being half a period long does not make an interval work. On this one $g$ rises and then falls, so some outputs repeat. Which half period matters, not just its length.
• D. The interval is right, but the reason is not. Many intervals half a period long fail, such as $0\le x\le \pi$. This one works because $g$ goes from its minimum to its maximum without turning back.