# Show that $f(x)=\int_0^{x^4}\sqrt{\frac{\cosh t}{1+\sqrt t}}dt$ has a local minimum at $x=0$

Let $f:\mathbb R\to\mathbb R$ define by $$f(x)=\int_0^{x^4}\sqrt{\frac{\cosh t}{1+\sqrt t}}dt,$$

is there an easy way to show that $f$ has a local minimum at $x=0$ ? I compute $f'(0)$, $f''(0)$ but both are $0$, and after it's complicate to compute $f'''$ and $f^{(4)}$. So may be there is an other way to get the result.

hint: use Taylor polynomial.

• The integrand is positive, so for all $x\neq 0$, the integral is positive. Therefore $0$ is a global minimum. – Git Gud Mar 18 '15 at 23:45
• Thanks for your answer. I forgot to precise that I have to use Taylor Polynomial. – ZiK789 Mar 19 '15 at 0:00
• Judging by the phrasing, you don't have to use Taylor's Polynomial. A hint is just a hint, you don't have to use it. – Git Gud Mar 19 '15 at 0:01

You can conclude with your method but as you remarked, it's a long way to conclude. Let's use Taylor polynomial. You can actually see that $$f'(x)=4x^3\sqrt{\frac{\cosh(x^4)}{1+\sqrt x^4}}dt.$$ Moreover, $$\sqrt{\frac{\cosh(0)}{1+\sqrt 0}}=1$$ and thus $$\sqrt{\frac{\cosh t}{1+\sqrt t}}=1+o(1)\quad \text{ if }\ t\to 0.$$
Therefore $$f'(x)=4x^3+o(x^3)$$ and thus $$f'(0)=f''(0)=f'''(0)=0$$ and $$\frac{f^{(4)}(0)}{3!}=4\implies f^{(4)}(0)=24>0.$$ You can finally conclude that $f$ has a local minimum at $x=0$.
I would think the easy way is to note that $f(0)=0$ and $f(x)>0$ if $x\not=0$ using that $\cosh(x)>0$ for all $x$ and the function under the integral sign is always positive. (You do not need any derivatives for this. You do not need Taylor polynomial either, though I used it, you just didn't see ... and I just saw I only repeated the comment by Git Gud that preceded both posted answers, sorry, I don't feel like erasing my answer now).