Find the minimum and maximum values of f(x) = (sin

Find the minimum and maximum values of $\displaystyle f(x) = \frac{\sin x-1}{\sin x-\cos x}$ on $\bigl[0,2\pi\bigr]$.

Step 1: Where the function is defined

Again $\tan x = 1$ makes the denominator vanish, so $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$ are excluded and the interval splits into three pieces.

Step 2: The derivative

With $u = \sin x-1$, $v = \sin x-\cos x$, the numerator of $f'$ is

\[ \cos x\bigl(\sin x-\cos x\bigr)-\bigl(\sin x-1\bigr)\bigl(\cos x+\sin x\bigr) = -\bigl(\sin^{2}x+\cos^{2}x\bigr)+\sin x+\cos x = \sin x+\cos x-1, \]

\[ f'(x) = \frac{\sqrt2\sin\left(x+\dfrac{\pi}{4}\right)-1}{\bigl(\sin x-\cos x\bigr)^{2}}. \tag{1} \]

Note this is exactly the negative of the numerator in Problem 444 — the two functions turn in opposite directions.

Step 3: Critical points and signs

The numerator is positive when $\sin\bigl(x+\tfrac{\pi}{4}\bigr)\gt \tfrac{1}{\sqrt2}$, i.e. on $\bigl(0,\tfrac{\pi}{2}\bigr)$, and negative on $\bigl(\tfrac{\pi}{2},2\pi\bigr)$. It vanishes at $x = 0$, $x = \dfrac{\pi}{2}$ and $x = 2\pi$, of which only $\dfrac{\pi}{2}$ is interior. There $f'$ passes from $+$ to $-$: a local maximum, with

\[ f\!\left(\frac{\pi}{2}\right) = \frac{1-1}{1-0} = 0. \]

Step 4: The poles

\[ \lim_{x\to(\pi/4)^{-}}f = +\infty, \qquad \lim_{x\to(\pi/4)^{+}}f = -\infty, \]

\[ \lim_{x\to(5\pi/4)^{-}}f = -\infty, \qquad \lim_{x\to(5\pi/4)^{+}}f = +\infty. \]

The function is unbounded in both directions: no global extrema exist.

Step 5: The four pieces

\[ \left[0,\tfrac{\pi}{4}\right):\ 1\nearrow+\infty; \qquad \left(\tfrac{\pi}{4},\tfrac{\pi}{2}\right]:\ -\infty\nearrow0; \]

\[ \left[\tfrac{\pi}{2},\tfrac{5\pi}{4}\right):\ 0\searrow-\infty; \qquad \left(\tfrac{5\pi}{4},2\pi\right]:\ +\infty\searrow1. \]

Useful values: $f(0) = 1$, $f(\pi) = -1$, $f\!\left(\tfrac{3\pi}{2}\right) = 2$, $f(2\pi) = 1$.

Step 6: Verification

\[f(1) = -0.5263795 \lt f\!\left(\tfrac{\pi}{2}\right) = 0 \gt f(2) = -0.0684318 \;\;\checkmark\ \text{(a maximum)}\]

\[f(0.1) = 1.0055809,\ \text{rising from } f(0) = 1 \text{ towards } +\infty \;\;\checkmark\]

\[f(6) = 1.0321315,\ \text{falling from } +\infty \text{ towards } f(2\pi) = 1 \;\;\checkmark\]