0
$\begingroup$

My attempt. I used the identity $\arctan(\alpha)\pm\arctan(\beta)=\arctan\left(\frac{\alpha\pm \beta}{1\mp \alpha\beta}\right)$ to rewrite $\arctan\left(\frac{x}{1+x^2}\right)$ as a sum of two $\arctan$'s, but I found that it did not help much with the sketch. Then I have been stuck until now. Any hint would be appreciated. Thank you!


Remark.
In case that someone wants to sketch some similar but more challenging function, the last part of the original question is to sketch $h(x)$ where $\tan(h(x))=\frac {x}{1-x^2}$, $-\infty\lt x\lt\infty$, and $h(0)=\pi$.
Last, this problem has been solved, but any new approach is always welcome. Thanks!

$\endgroup$
3
  • 1
    $\begingroup$ Can you not differentiate $tan^{-1}(\frac{x}{1+x^2})$? $\endgroup$
    – DatBoi
    Aug 28, 2020 at 7:23
  • $\begingroup$ Hi, DatBoi. Sorry for replying late. I have no idea why I did not think about taking the derivative. Thanks a lot! @DatBoi $\endgroup$ Aug 28, 2020 at 10:54
  • 1
    $\begingroup$ That's ok. I'd advise you to stay active for 1-2 hours after posting a question. You are likely to get a reply. Also note that I have not used double derivative. A bit of logic can be used to identify possible curves and eliminating some with already known equations. This is particularly helpful in objective exams $\endgroup$
    – DatBoi
    Aug 28, 2020 at 12:31

2 Answers 2

1
$\begingroup$

$$\frac{d(\arctan(\frac{x}{1+x^2}))}{dx}=\frac{1}{1+\frac{x^2}{(1+x^2)^2}}.\frac{1+x^2-2x^2}{(1+x^2)^2}$$ $$f'(x)=\frac{1-x^2}{x^4+3x^2+1}$$ $$f'(x)=0,x=\pm1$$ Also $f(0)=0$

and $\lim_{x\to\pm\infty}{f(x)}=0$

Therefore the graph must be one of these:

But we know that $f(x)>0$ when $x>0$ and $f(x)<0$ when $x<0$ So it is the first one.

$\endgroup$
1
$\begingroup$

$g(x)=\tan^{-1} \frac{x}{1+x^2}$, $g(0)=0$, $g(x)$ is an odd function. $g(\pm \infty)=0$ $$g'(x)=\frac{1-x^2}{x^4+3x^2+1}=0\implies x=\pm 1 $$ $g'(x)$ changes sign from negative to positive when we pass $x=-1$ so there is minimu and similarly max at $x=1$. Hence $f_{max}=f(1)=\tan^{-1}(1/2)=0.4634, f_{min}=-0.4634.$ Therefore $g(x)$ a curve which is continuous and odd. It passes through origin vanishes asymptotically and has a max and min at $x=\pm 1.$ See the sketch of $g(x)$ in the fig. below $g(x)$

enter image description here

$\endgroup$

You must log in to answer this question.

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