0
$\begingroup$

We encountered this question in the class today. My lecturer asked me to define a function $f(x)=\tan(x) - x$. Then we were asked to find its derivative, $f'(x)$, i.e, $\sec^2(x) - 1$, and solve the inequality $f'(x)>0$. The answer was $(0,π/2)$.

My question is, if $f'(x)>0$, it just proves that $f(x)$ is increasing in the interval, it doesn't prove if $f(x)$ is always positive there. Ok, agreed that since $\tan0=0$, $\tan x>x$ in $(0,π/2)$ since it's increasing. But there is possibility of loosing a solution right? It could be possible that in an interval $\tan x-x$ is decreasing but $\tan(x)$ is still $>x$.

I'm confused.

$\endgroup$
3
  • $\begingroup$ Why is the answer just $(0,\pi/2)$ when $\sec^2-1$ is positive everywhere except at integer multiples of $\pi/2$? Was the domain of $f$ already defined to be $[0,\pi/2)$? $\endgroup$
    – David K
    Jun 17 '18 at 19:30
  • $\begingroup$ We had options 1) (0,π/2) 2) (π/2, π) 3)(-π/2, 0) 4) (-π/2,π/2) $\endgroup$
    – Dante
    Jun 17 '18 at 19:43
  • $\begingroup$ It seems to me that all options except number 4 are subsets of the solution set, unless some other conditions were stated that would rule out 2 and 3. $\endgroup$
    – David K
    Jun 17 '18 at 20:34
1
$\begingroup$

$f'(x)>0$ in the interval $(0, \frac{\pi}{2})$, so $f$ is increasing on this interval. Since $f(0)=0$ and $f$ is continuous on the interval, $f$ must be positive on the interval. Note that $f(\frac{\pi}{2})$ is undefined, and then $f(\frac{\pi}{2}+ \epsilon)$ for some small $\epsilon>0$ is negative. So we can say that $f$ is always positive on the interval you've specified.

$\endgroup$
0
$\begingroup$

Note that the derivative is positive everywhere it is defined, i.e. for all $x\ne(2k+1)\dfrac\pi2$.

For a continuous function and $x_1>x_0$, if you know $f(x_0)>0$ and $f$ increasing, then $f(x_1)>0$. But if $f(x_0)>0$ and $f$ decreasing, you can't conclude about the sign of $f(x_1)$.

$\endgroup$
0
$\begingroup$

No. $\tan x -x$ is always increasing on each interval on which it is defined, since it is differentiable, and its derivative has a second form: $$(\tan x-x)'=1+\tan^2x-1=\tan^2x>0$$ except at the isolated points $\;\Bigl\{\dfrac\pi2+k\pi\:\Big\vert\: k\in\mathbf Z\Bigr\}$, where it is equal to $0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.