Science:Math Exam Resources/Courses/MATH100/December 2016/Question 14 (b)

MATH100 December 2016

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Question 14 (b)
Let $f(x)$ be a function so that $f(x),f'(x)$, $f''(x)$ exist and are continuous for all $x$, and $|f(x)-\sin x|\leq 1/3$ for all $x$. Show that $f''(x)$ has at least one zero in the open interval $(0,2\pi )$.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Using the mean value theorem, show that the second derivative of $f$ must be negative somewhere between $0$ and $\pi$. Also, the second derivative must be positive somewhere between $\pi$ and $2\pi$.

Hint 2
Apply the intermediate value theorem to the second derivative of $f$ and find a zero of $f''(x)$.

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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. In the similar way in part (a), we observe that $f(0)\leq {\frac {1}{3}}$ $f\left({\frac {\pi }{2}}\right)\geq {\frac {2}{3}}$ $-{\frac {1}{3}}\leq f(\pi )\leq {\frac {1}{3}}$ $f\left({\frac {3\pi }{2}}\right)\leq -{\frac {2}{3}}$ $f(2\pi )\geq -{\frac {1}{3}}$ This implies that the derivative must be positive somewhere between $0$ and ${\frac {\pi }{2}}$ and then negative somewhere between ${\frac {\pi }{2}}$ and $\pi$. This is because by the mean value theorem, we have $c_{1}\in \left(0,{\frac {\pi }{2}}\right)$ and $c_{2}\in \left({\frac {\pi }{2}},\pi \right)$ such that $f'(c_{1})={\frac {f\left({\frac {\pi }{2}}\right)-f(0)}{\frac {\pi }{2}}}\geq {\frac {{\frac {2}{3}}-{\frac {1}{3}}}{\frac {\pi }{2}}}={\frac {2}{\pi }}\cdot {\frac {1}{3}}>0$ and $f'(c_{2})={\frac {f(\pi )-f\left({\frac {\pi }{2}}\right)}{\frac {\pi }{2}}}\leq {\frac {{\frac {1}{3}}-{\frac {2}{3}}}{\frac {\pi }{2}}}=-{\frac {2}{\pi }}\cdot {\frac {1}{3}}<0.$ Thus the second derivative must be negative somewhere between $0$ and $\pi$; we can find $c_{3}\in (0,\pi )$ such that $f''(c_{3})={\frac {f'(c_{2})-f'(c_{1})}{c_{2}-c_{1}}}<0.$ Similarly, the derivative must be negative somewhere between $\pi$ and $3\pi /2$ and then positive between $3\pi /2$ and $2\pi$. Hence the second derivative must be positive somewhere between $\pi$ and $2\pi$. Thus, by the intermediate value theorem, the second derivative has a zero (and changes sign at that zero) somewhere between $0$ and $2\pi$.