Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0$

Question

Find the Maximum and minimum values of $f(x)=\frac{\cos x- \sin x}{\cos x+\sin x}, x\geq0 $ tanx$ \neq -1$

I've tried to find the derivative: $$ f'(x)= \frac{-(\cos x+\sin x)(\sin x+\cos x)-(\cos x-\sin x)^2}{(\cos x+\sin x)^2}$$ $$f'(x)=\frac{-2}{ (\cos x+\sin x)^2 }$$

$f'(x)<0$ as tan x $\neq$ -1

$x\geq 0 \Rightarrow f(x)<f(0)$

$f(x)<1$.

So maximum value is 1. Is this correct? What is the minimum value?

Answer 1

Hint: Divide by $\cos x$.

$$\dfrac{\cos x-\sin x}{\cos x+\sin x}$$ $$=\dfrac{1-\tan x}{1+\tan x}$$ $$=\dfrac{\tan\left(\dfrac{\pi}{4}\right)-\tan x}{1+\tan\left(\dfrac{\pi}{4}\right)\cdot\tan x}$$ $$=\tan\left(\frac{\pi}{4}-x\right)$$

Therefore the range of this function extends from $-\infty$ to $+\infty$.

Answer 2

I would try it another way:

$$ \cos(x)-\sin(x) = \sqrt{2}\cos(x+\frac{\pi}{4}) \\ \cos(x)+\sin(x) = \sqrt{2}\cos(x-\frac{\pi}{4}) $$

then

$$ f(x)=\frac{\cos(x+\frac{\pi}{4})}{\cos(x-\frac{\pi}{4})}=-\frac{\sin(x-\frac \pi 4)}{\cos(x-\frac \pi 4)}=-\tan(x-\pi/4) \\ $$

As you might know, the function $\tan(x)$ is not bounded. Hence your $f(x)$ has no global maxima or minima. E.g. it takes all values from $-\infty$ to $\infty$.