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

MATH101 April 2018

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Question 04 (ii)
Define ${\textstyle F(x)}$ and ${\textstyle g(x)}$ by ${\textstyle F(x)=\int _{0}^{x}(t^{4}+1)\sin {t}\,dt}$ and ${\textstyle g(x)=F(x^{2})}$ . Determine the smallest value of ${\textstyle x>0}$ at which ${\textstyle g(x)}$ has a local minimum.

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Hint 1
If a differentiable function $g$ satisfies $g'(c)=0$ and $g'$ changes its sign from minus to plus at $x=c$ , $g$ has its local minimum at $x=c$ .

Hint 2
To find the derivative of $g$ , use the chain rule and the fundamental theorem of Calculus.

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Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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: $\color {blue}{\sqrt {2\pi }}$