0
$\begingroup$

How to show that $$f(x)=1+|\cos^n(x)|$$ is monotonic function for $x>0$?

$n$ is fixed here. How to take derivative here?

Is there an other way to prove it?

Thanks

$\endgroup$
  • $\begingroup$ What is $n$ in this case? $\endgroup$ – Vasting Apr 8 '19 at 19:25
  • $\begingroup$ What can you conclude from wolframalpha.com/input/?i=plot+1+%2B+abs(cos(x)%5E2) ? $\endgroup$ – Martin R Apr 8 '19 at 19:27
  • $\begingroup$ $n \mapsto 1 + |\cos^n (x)|$ is certainly monotonic decreasing in $n$ for a fixed real value of $x.$ $\endgroup$ – Dbchatto67 Apr 8 '19 at 19:30
6
$\begingroup$

Intuitive answer: It is not monotonic, because however far you go along the $x$-axis, there will always be infinitely many values of $x$ such that $|\cos^n x|=1$ and values of $x$ such that $|\cos^nx|=0$.

The derivative of this function is defined everywhere and equal to $-\text{sgn}(\cos^nx) \cdot n\cos^{n-1}x\sin x$ except for maybe in the points $x=(k+\frac{1}{2})\pi$. In those points, you have to check for differentiability by definition.

However, you don't even need to know the derivative at those special points. It suffices to see that in some points of differentiability $f'(x)>0$ and at some points of differentiability $f'(x)<0$. This means that on some parts the function is increasing and on some parts it is decreasing, hence the function is not monotonic.

| cite | improve this answer | |
$\endgroup$

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.