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 $$F^{\prime }(x)=\frac{d}{dx}(\underset{0}{\overset{x}{\int }}(t^4+1)\sin tdt)=(x^4+1)\sin x.$$

Then, the derivative of $g$ is

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

Since $x^8+1\ge 1$ for any $x$ on the real line and we consider only positive $x$, the sign of $g^{\prime }$ 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,

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

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 $g^{\prime }$. Therefore, the smallest number $x$ at which $g$ has a local minimum is $x=\sqrt{2\pi }$.

Answer: $\sqrt{2\pi }$