Sketch $f(x)=\sin x+\frac{1}{x}$ finding local maxima and minima, intervals of increase and decrease. I'm trying to use differentiation to draw this picture and find critical points.

So, I get $f'(x)=\cos x-\frac{1}{x^2}$ However, I'll have to deal with inequality $f'(x)>0 $, $f'(x)<0$, and $f'(x)=0$, I feel I lack an ability to solve an equation like this. So is there any better way to find local maximun and minumum for this function?


1 Answer 1


For $f'$, note that, if $|x| < 1$, then $f'(x) \ne 0$ (do you see why?).

Also, for large $x$, since $\cos(x) =\sin(\pi/2-x) $, you want $\sin(\pi/2-x)$ small. Therefore, $\pi/2-x$ has to be close to a multiple of $\pi$, since that is when $\sin$ is small. So, let $\pi/2 -x=\pi n +z$, where $z$ is small. Then $\cos(x) =\sin(\pi/2-x) =\sin(\pi n+z) \approx (-1)^n z $, so $\frac1{x^2} \approx (-1)^n z \approx (-1)^n (\pi/2-\pi n)-x $.

Since $x$ is large, $x \approx (-1)^n (\pi/2-\pi n) $.

Plot $f$ and $f'$ and see if these remarks agree with what you see.

Remember, the most useful approximation for trig functions is $\sin(x) \approx x$ for small $x$.

  • $\begingroup$ I understand how you get $x$ for $f'(x)=0$ But I still don't know how to calculate the range of an inequality as $f'(x)>0$ and $f'(x)<0$ Can you show me how to get the interval? $\endgroup$
    Nov 29, 2015 at 23:16

You must log in to answer this question.

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