Science:Math Exam Resources/Courses/MATH102/December 2014/Question B 01/Solution 1

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