Find all critical points and say are they extremum points or not.
$$f(x)=\cos x +\cosh x$$
$$\frac{df}{dx}=\sinh x-\sin x$$
$\sinh x-\sin x=0$ for $x=0$ is true but how prove there are no other roots?
$\dfrac{d^2f}{dx^2}=\cosh x-\cos x$ if we proved that $x=0$ is only root then second question is see where second derivative is positive and negative but I can't do it can you help for this two questions?
Best Answer
If $x>0$, then\begin{align}\sinh(x)&=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+\cdots\\[6pt]&>x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-\cdots\\[6pt]&=\sin(x).\end{align}And, if $x<0$, $\sinh(x)=-\sinh(-x)<-\sin(-x)=\sin(x)$. So, you only have $\sinh(x)=\sin(x)$ when $x=0$.
Now, it follows from the fact that $f'(0)=f''(0)=f'''(0)=0$ and that $f^{(4)}(0)=2>0$ that $f$ has a minimum at $0$.