Science:Math Exam Resources/Courses/MATH102/December 2016/Question A 08

MATH102 December 2016

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Question A 08

Which of the following statements is correct? (i) Newton’s method finds the zero closest to the initial guess $x_{0}$. (ii) The function $\cos(x)$ has a tangent line at some $x\in [-\pi ,\pi ]$ that goes through the point $(0,2)$. (iii) A tangent line to the graph of $f(x)$ at an inflection point must cross from one side to the other side of the graph of $f(x)$. (iv) If neither $f(x)$ nor $g(x)$ are differentiable at $x=a$ then the function $h(x)=f(x)+g(x)$ cannot be differentiable at $a$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The tangent line of $f(x)$ at the point $x=a$ is $y=f'(a)(x-a)+f(a)$.

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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. The correct choice is (iii), because at the inflection point the function changes its concavity, so the tangent line moves from being above the graph to be below the graph or vice versa, so it must cross the graph from one side to the other. (i) is not correct. See the Questions B 01; Although the zero $z_{2}$ is closer than another zero $z_{1}$ to the initial point $B$, but the newton method starting at $B$ finds the zero $z_{1}$. (ii) is not correct. The tangent line of $f(x)=\cos x$ at a point $a$ is $y=f'(a)(x-a)+f(a)=-\sin a(x-a)+\cos a$. Assume that there is a tangent line which goes through $(0,2)$. This means that there exists a point $a\in [-\pi ,\pi ]$ such that $2=-\sin a(0-a)+\cos a=a\sin a+\cos a$. However there's no such point $a\in [-\pi ,\pi ]$ because the maximum of $a\sin a+\cos a$ is ${\frac {\pi }{2}}<2$. This is a contradiction. We find the maximum in the following way. Let $g(a)=a\sin a+\cos a$. Then, its derivative is $g'(a)=a\cos a$, which vanishes at $a=0,\pm {\frac {\pi }{2}}$. Indeed, these are critical points. so that the maximum in $[-\pi ,\pi ]$ is achieved either the critical points $0,\pm {\frac {\pi }{2}}$ or the end points $\pm \pi$. Since $g(0)=0,g\left({\frac {\pi }{2}}\right)={\frac {\pi }{2}},g(\pm \pi )=-1$, the maximum is ${\frac {\pi }{2}}$. (iv) is not correct because we can take $f(x)=|x|$, $g(x)=-|x|$, and $a=0$; Obviously, $f$ and $g$ are NOT differentiable at $x=a=0$. but $h(x)=f(x)+g(x)=|x|-|x|=0$ is a constant function and hence is differentiable at $x=a$. (Proof for (iii)) We need to show $g(x)=f(x)-({\text{tangent line}})=f(x)-f'(a)(x-a)-f(a)$ changes the sign around $x=a$. Since $g(a)=0$, it is enough to show that $g$ is either increasing or decreasing around $x=a$. (If it is increasing the sign is changing from negative to positive, otherwise it is from positive to negative.) For this purpose, we calculate the derivative $g'(x)=f'(x)-f'(a)=\int _{a}^{x}f''(t)dt$. Since $f''$ changes the sign around $x=a$, $g'$ doesn't change its sign around $x=a$. For example, in the case that $f''$ change its sign from negative to positive around $x=a$, then for $x<a$ sufficiently close with $a$, $g'(x)=\int _{a}^{x}f''(t)dt=-\int _{x}^{a}f''(t)dt>0$ and for $x<a$ sufficiently close with $a$, $g'(x)=\int _{a}^{x}f''(t)dt>0$. Therefore, we prove that $g$ is either increasing or decreasing around $x=a$. Answer: $\color {blue}(iii)$