Show the following function has a local minima in $0$:

Question

Given the function $f:\mathbb{R}\longrightarrow\mathbb{R}$ defined as:

$$f(t)=\int_0^t\sin(t\sqrt{1+x^2})dx$$

show $f$ has a local minima in $0$. My question is: can we show that $f$ has a local minimum in $0$ using its asymptotic behaviour?

Indeed, we can see that $f(t)\sim t\int_0^t\sqrt{1+x^2}dx=g(t)$ when $|t|<1$, thus $f'(t)\sim g'(t)=\int_0^t\sqrt{1+x^2}dx+t\sqrt{1+t^2}$ and we have that $g'(0)=0$, $g'(t)>0$ when $t>0$ and $g'(t)<0$ when $t<0$ (provided we restrain ourselves at $|t|<1$) hence concluding the exercise.

Answer 1

There is no need to compute the derivative since the sign of $f(t)$ can be easily determined for small $t$:

• $f(0) = 0$.

• For $0 < t < 1$ and $0 \le x \le t$ is $$ 0 < t\sqrt{1+x^2} \le \sqrt 2 \implies \sin(t\sqrt{1+x^2}) > 0 $$ and therefore $f(t) > 0$.

• Similarly, $f(t) > 0$ for $-1 < t < 0$.

It follows that $f$ has a local minimum at $t=0$.