Show $f(x)=x\sec(x)$ is one-to-one on$\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$

Question

So as the title states I have to show that the above function $f(x)=x\sec(x)$ is one-to-one on the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

However, i'm not entirely sure how without testing every number between the given interval?

Any suggestion would be appreciated.

Answer 1

Computing the derivative of your function, we find $$ f'(x)=\frac{\cos x+x\sin x}{\cos^2 x} $$ which vanishes whenever $$ -x=\tan x $$ which by inspection occurs when $x=0$, and nowhere else, since $$ (\tan x+x)'=\sec^2x+1>0 $$ So we test on either side of $0$, the only place we could have the problem of $f'$ changing signs, for positive $x$, that $f'>0$ is clear. We check the convenient value $x=-\pi/4$ on the left to find $$ \cos(\pi/4)-\pi/4\sin(\pi/4)=\sqrt{2}/2(1-\frac{\pi}{4})>0 $$ so yes, it is indeed invertible.

Answer 2

$\frac{d}{dx} x\sec(x) = (x\tan(x)+1)\sec(x) > 0 $ on the given interval.

Answer 3

$f(x)=\frac{x}{\cos x}$ is an odd function and $\lim_{x\to \left(\pm\frac{\pi}{2}\right)^{\mp}}f(x)=\pm\infty$, $\lim_{x\to 0}f(x)=0$. $f(x)$ is positive on $\left(0,\frac{\pi}{2}\right)$ and over such interval $\log f(x) = \log(x)-\log(\cos x)$ is increasing, since $\frac{d}{dx}\log f(x)$ equals $\frac{1}{x}+\tan(x)>0$. It follows that $f(x)$ is increasing over $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Answer 4

Because $\cos x$ is strictly decreasing and nonnegative on $[0,\pi/2),$ $\sec x = 1/\cos x$ is strictly increasing and nonnegative on $[0,\pi/2).$ The function $x$ is also strictly increasing and nonnegative on $[0,\pi/2).$ Therefore the product $x\sec x$ is strictly increasing and nonnegative on $[0,\pi/2).$

Now any odd fuction on $(-\pi/2,\pi/2)$ that is strictly increasing on $[0,\pi/2)$ is strictly increasing on $(-\pi/2,\pi/2).$ Since $x\sec x$ is odd, we have the desired result.