How can I show a decreasing derivative?

Question

I have a derivative $f'(x) = -1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x}$ And we have to show that it's decreasing in all intervals on the form $[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}$ for n>= 1 Our hint is to use the derivative such as $\frac {6}{(12n + 7)\pi} <= \frac{6}{19 \pi} < \frac{1}{8}$ I have no idea what to do other than $f'(x) = - 1 + 8 \cos \frac{1}{x} + 4 \sin \frac{1}{x} < \frac {6}{(12 n + 11) \pi } $ Please help. And thank you.

Answer 1

Hint: $$x \in \left[\frac {6}{(12 n + 11) \pi } , \frac {6}{(12n + 7)\pi}\right] \iff \frac1x \in \left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$$

We have $x,y \in \left[\frac {6}{(12 n + 11) \pi },\frac {6}{(12n + 7)\pi}\right]$ and $x<y$ iff $\frac1x>\frac1y$ in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$. So this question can be rewritten as proving that $g(x) = 1+8\cos x+4\sin x$ is increasing in $\left[\left(2n+\frac76\right)\pi,\left(2n+\frac{11}6\right)\pi\right]$. Since $\sin$ and $\cos$ are periodic with period $2\pi$, the "$2n$"'s in the previous interval are redundant. You may continue from here. By either calculating $g'(x)$ OR rewriting the two trigonometric terms as one single term with a phase change.