Science:Math Exam Resources/Courses/MATH102/December 2013/Question C 03/Solution 1

We begin by finding the zeroes of the original function,

$\displaystyle 0=x^{2}e^{-x}$

and since $\displaystyle e^{-x}>0$ always, only $\displaystyle x=0$ is a root. But we note that $f$ is always greater or equal to zero.

Next we take the derivative and see that

$\displaystyle f'(x)=2xe^{-x}-x^{2}e^{-x}=xe^{-x}(2-x)$

Finding the roots gives us $\displaystyle x=0,2$. Making a sign chart, we see that

1. The function is decreasing between $\displaystyle (-\infty ,0)$
2. The function is increasing between $\displaystyle (0,2)$
3. The function is decreasing between $\displaystyle (2,\infty )$

Graphically, we have

and thus we have that $x=0$ is a local minimum and $x=2$ is a local maximum. The coordinates are given by $(0,0)$ and $(2,4e^{-2})$ (note that $4e^{-2}$ is roughly $4/9$).

Lastly taking the second derivative, we have that

$\displaystyle f''(x)=2(e^{-x}-xe^{-x})-(2xe^{-x}-x^{2}e^{-x})=e^{-x}(2-2x-2x+x^{2})=e^{-x}(2-4x+x^{2})$

The zeroes of the second derivative are given by $\displaystyle x^{2}-4x+2=0$.

Using the quadratic formula gives us the roots

$x={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(2)}}}{2}}={\frac {4\pm {\sqrt {8}}}{2}}={\frac {4\pm 2{\sqrt {2}}}{2}}=2\pm {\sqrt {2}}$

Making a sign chart, we see that

1. The function is concave up between $\displaystyle (-\infty ,2-{\sqrt {2}})$
2. The function is concave down between $\displaystyle (2-{\sqrt {2}},2+{\sqrt {2}})$
3. The function is concave up between $\displaystyle (2+{\sqrt {2}},\infty )$

Graphically we have

Hence, we have that $2\pm {\sqrt {2}}$ are inflection points. Combining all this information, we have the following. First plot all known points of interest and label where the function is increasing, decreasing, concave up and concave down. Then connect the dots in a consistent way. Don't forget that $\lim _{x\to \infty }x^{2}e^{-x}=0$ was given to us and we can use this information.