Science:Math Exam Resources/Courses/MATH101/April 2018/Question 04 (ii)/Solution 1

To find the positive points at which local minimums of $g$ occur, let's find the derivative of $g$ first.

Since $g$ can be written as a composite of two functions, $g(x)=F(x^{2})=F(h(x))$, where $h(x)=x^{2}$, we can apply the chain rule to find its derivative. In this process, we need the derivative of $F$.

By the Fundamental theorem of Calculus, we have $$F'(x)={\frac {d}{dx}}\left(\int _{0}^{x}(t^{4}+1)\sin {t}dt\right)=(x^{4}+1)\sin x.$$

Then, the derivative of ${\textstyle g}$ is

$$g'(x)=(F(h(x))'=F'(h(x))h'(x)=F'(x^{2})(x^{2})'=2x(x^{8}+1)\sin(x^{2}).$$

Since $x^{8}+1\geq 1$ for any $x$ on the real line and we consider only positive $x$, the sign of $g'$ is determined by $\sin(x^{2})$.

Note that $\sin(x^{2})=0$ at $x^{2}=\pi ,2\pi ,\cdots$ (i.e., $x={\sqrt {\pi }},{\sqrt {2\pi }},\cdots$). Also,

$${\begin{aligned}&\sin(x^{2})>0,\quad {\text{on }}(0,{\sqrt {\pi }}),({\sqrt {2\pi }},{\sqrt {3\pi }}),\cdots \\&\sin(x^{2})<0,\quad {\text{on }}({\sqrt {\pi }},{\sqrt {2\pi }}),({\sqrt {3\pi }},{\sqrt {4\pi }}),\cdots .\end{aligned}}$$

Since the sign of $\sin(x^{2})$ is changed from minus to plus at $x={\sqrt {2\pi }}$ for the first time among positive $x$, so is the sign of $g'$. Therefore, the smallest number $x$ at which $g$ has a local minimum is $x={\sqrt {2\pi }}$.

Answer: $\color {blue}{\sqrt {2\pi }}$