0
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ Multiplying (or dividing) by a negative number reverses the sign of an inequality. :) $\endgroup$ Commented Jun 28, 2014 at 22:43
  • 1
    $\begingroup$ When you divide/multiply an inequality by a negative number, the inequality signs need to be flipped. $\endgroup$ Commented Jun 28, 2014 at 22:43
  • $\begingroup$ I did that didnt I? That was the first thing that came to my mind, maybe I overlooked it. Could you please post a correct answer and highlight where I've gone wrong. Thanks! $\endgroup$
    – seeker
    Commented Jun 28, 2014 at 22:44

1 Answer 1

5
$\begingroup$

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

$$-\sec x\ge1\gt0$$

$\endgroup$
4
  • $\begingroup$ OH! Thank you so much! $\endgroup$
    – seeker
    Commented Jun 28, 2014 at 22:46
  • $\begingroup$ quick question; I know the $\cos (x)$ in the inequality $-1\leqslant\cos(x)<0$ is negative. But why does that make the $\cos(x)$ I am dividing through by negative? Am I not dividing through by a positive $\cos(x)$, which would not impact the inequality at all? $\endgroup$
    – seeker
    Commented Jun 29, 2014 at 11:28
  • 2
    $\begingroup$ @Assad, a good way to understand this is to explicitly include the logical connectives in what you're doing. What you're actually saying is, If $-1\le\cos x\lt0$, then ${-1\over\cos x}\ge{\cos x\over\cos x}\gt{0\over\cos x}$. The if part establishes that you're talking about values of $x$ for which $\cos x$ is negative. $\endgroup$ Commented Jun 29, 2014 at 14:16
  • $\begingroup$ I see, you've cleared it up, thanks. $\endgroup$
    – seeker
    Commented Jun 29, 2014 at 14:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .