Science:Math Exam Resources/Courses/MATH102/December 2014/Question B 01

MATH102 December 2014

• QA 1 • QA 2 • QA 3 • QA 4 • QA 5 • QA 6 • QA 7 • QA 8 • QB 1 • QB 2 • QB 3 • QB 4 • QB 5 • QB 6 • QB 7 • QC 1 • QC 2(a) • QC 2(b) • QC 2(c) • QC 2(d) • QC 3 •

Other MATH102 Exams

• December 2014 • December 2011 • December 2012 • December 2013 • December 2016 • December 2015 •

[hide] Question B 01
List the x-coordinates of all the minima, maxima and inflection points of the function $f(x)=e^{-x^{2}+x}.$

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 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.

[show] Hint 1
Use first derivative to identify critical points.

[show] Hint 2
Use second derivative test to verify whether the obtained critical points are local maxima or minima and also to find inflection points.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.

If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

[show] Solution
First we calculate $f'(x)$ by the chain rule. We write the function as $\displaystyle g(h(x))$ where $\displaystyle g(x)=e^{x}$ and $\displaystyle h(x)=-x^{2}+x$ . The chain rule states that $\displaystyle {\frac {d}{dx}}g(h(x))=g'(h(x))h'(x)$ and thus $f'(x)=(-2x+1)e^{-x^{2}+x}.$ Since $e^{y}>0$ for any real number $y$ , $x=1/2$ is the only critical point. i.e., $f'(x)=0\implies x={\frac {1}{2}}$ . To determine whether it is a maximum or minimum, we calculate the second derivative using a combination of the product and chain rule. This gives ${\begin{aligned}f''(x)&=\left({\frac {d}{dx}}(-2x+1)\right)\cdot e^{-x^{2}+x}+(-2x+1){\frac {d}{dx}}e^{-x^{2}+x}\\&=-2e^{-x^{2}+x}+(-2x+1)(-2x+1)e^{-x^{2}+x}\\&=-2e^{-x^{2}+x}+(-2x+1)^{2}e^{-x^{2}+x}\\&=(-2+(2x-1)^{2})e^{-x^{2}+x}\end{aligned}}$ As $f''({\frac {1}{2}})<0$ , we know that $x={\frac {1}{2}}$ is a max. To find the inflection points, we solve $f''(x)=0$ . As exponentials are always positive, these occur when $\displaystyle -2+(-2x+1)^{2}=0$ or, solving for $x$ , when $x_{1}={\frac {1-{\sqrt {2}}}{2}},\quad x_{2}={\frac {1+{\sqrt {2}}}{2}}.$ Notice the two points divide the whole domain into three intervals and in each of those intervals, plugging in $-10$ , $0$ and $10$ , we have ${\begin{aligned}f''(x)>0,&\quad x<x_{1},\\f''(x)<0,&\quad x_{1}<x<x_{2},\\f''(x)>0,&\quad x_{2}<x,\end{aligned}}$ So $f''(x)$ change signs on both sides of $x_{1}$ and $x_{2}$ . So $x_{1},x_{2}$ are inflection points. Answer: The given function $f$ one point $x_{max}$ having its maximum and two inflection points $x_{inf}^{1}$ and $x_{inf}^{2}$ ; $\color {blue}x_{max}={\frac {1}{2}},x_{inf}^{1}={\frac {1-{\sqrt {2}}}{2}},x_{inf}^{2}={\frac {1+{\sqrt {2}}}{2}}$ .