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

MATH102 December 2014

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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.

Hint 1
Use first derivative to identify critical points.

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

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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. 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 $x_{inf}^{2}$; $\color {blue}x_{max}={\frac {1}{2}},x_{inf}^{1}={\frac {1-{\sqrt {2}}}{2}},x_{inf}^{2}={\frac {1+{\sqrt {2}}}{2}}$.

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