I'm currently attempting to find the range of $f(x)=\sec(x)$ by considering $\cos(x)$ in the intervals of $0<\cos(x)\leqslant 1$ and $-1\leqslant \cos(x)<0$ (as $\sec(x)$ is undefined for $\cos(x)=0$).
What I have here are two compound inequalities, if I divide through by $\cos(x)$, i'll end up with an inequality in terms of $\sec(x)$; which essentially what amounts to the range of $f(x)=\sec(x)$.
For the first inequality I get:
$\frac{0}{\cos (x)}<\frac{\cos(x)}{\cos(x)}\leqslant \frac{1}{\cos(x)}$
Which essentially simplifies to: $1\leqslant\sec(x)$, therefore $\sec(x) \geqslant1$ This makes sense.
However for the second inequality I experience problems:
$-1\leqslant \cos(x)<0$
$\frac{-1}{\cos (x)}\leqslant \frac{\cos(x)}{\cos(x)}< \frac{0}{\cos(x)}$
$-\sec(x)\leqslant 1<0$
Which is absurd, as 1 is not less than zero.
I've tried to separate the two inequalities but I still don't get the answer I want:
$-1\leqslant \cos(x)<0$ separates to $-1\leqslant\cos(x)$ and $\cos(x)<0$
(I ignore the $\cos(x)<0$, as I dont believe that is pertinent in obtaining the range of $sec(x)$)
So carrying on with the inequality $-1\leqslant\cos(x)$ as it seems more promising, I get:
$\cos(x)\geqslant-1$
$\frac{\cos(x)}{\cos(x)}\geqslant\frac{-1}{\cos(x)}$
$1\geqslant-\sec(x)$
$-\sec(x)\leqslant1$
Dividing through by $-1$:
$\sec(x)\geqslant1$
Which is wrong, because in the interval where $-1\leqslant \cos(x)<0$, $\sec(x)\leqslant-1$
Any help would be appreciated, this is stressing me out. What am I doing wrong?
You are taking the second inequality, $-1\le\cos x\lt0$, and dividing everything by $\cos x$. But $\cos x$ is negative here, and when you divide (or multiply) both sides on an inequality by a negative number, the inequality sign reverses direction. So you should get
$${-1\over\cos x}\ge{\cos x\over\cos x}\gt{0\over\cos x}$$
in this case, which simplifies to
$$-\sec x\ge1\gt0$$