7.5.3 Exercises

Maximum Domain
We have $f(x)=-x(x-2)e{}^x$ and $e{}^x>0$ for all $x\in ℝ$; thus, $D_f=ℝ=]-\infty ;\infty [$ is the maximum domain.  
$x$- and $y$-Intercepts
The intersection points with the $x$-axis (roots of the function) lie at $x_1=0$ and $x_2=2$, i.e. the coordinates of the points are $(0;0)$ and $(2;0)$. The $y$-intercept is the point $(0;0)$.  
Symmetry
The function $f$ is neither even nor odd, and hence the graph of $f$ is neither axially symmetric with respect to the $y$-axis nor centrally symmetric with respect to the origin.  
Limiting Behaviour
Since the function is defined for all real numbers, only the asymptotes for $\pm \infty$ have to be investigated:

${\displaystyle \underset{x\to \infty }{lim}-x(x-2)e{}^x\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}-\infty \mathrm{\hspace{0.5em}\hspace{0.5em}}\text{and}\mathrm{\hspace{0.5em}\hspace{0.5em}}\underset{x\to -\infty }{lim}-x(x-2)e{}^x\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}0\hspace{0.5em}.}$

Hence, $y=0$ is an asymptote for $x\to -\infty$.  
Derivatives
The first two derivatives of $f$ are

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{f\text{'}(x)}& =\hfill & (2-2x)e{}^x+(2x-x^2)e{}^x=(2-x^2)e{}^x=-(x^2-2)e{}^x\hspace{0.5em},\hfill \\ \multicolumn{1}{c}{f\text{'}\text{'}(x)}& =\hfill & -2xe{}^x+(2-x^2)e{}^x=-(x^2+2x-2)e{}^x\hspace{0.5em}.\hfill \end{array}}$
 
Monotony Behaviour and Extremal Values
The solutions of $f\text{'}(x)=0$ are $x_1=-\sqrt{2}$ and $x_2=\sqrt{2}$. Furthermore, we have $x_1<x_2$ and

${\displaystyle f\text{'}(x)\mathrm{\hspace{0.5em}\hspace{0.5em}}=\mathrm{\hspace{0.5em}\hspace{0.5em}}-(x+\sqrt{2})(x-\sqrt{2})e{}^x\hspace{0.5em}.}$

On $]-\infty ;-\sqrt{2}[$ the first derivative $f\text{'}$ is negative, so $f$ is monotonically decreasing there. On $]-\sqrt{2};\sqrt{2}[$ the first derivative $f\text{'}$ is positive, hence $f$ is monotonically increasing there. On $]\sqrt{2};\infty [$ the first derivative $f\text{'}$ is negative, so $f$ is monotonically decreasing there. Thus, $x_1=-\sqrt{2}$ is a minimum point, and $x_2=\sqrt{2}$ is a maximum point.  
Inflexion Points
The necessary condition $f\text{'}\text{'}(x)=0$ for $x$ to be an inflexion point results in the quadratic equation $x^2+2x-2=0$. It has the solutions $w_1=\frac{-2-\sqrt{4+8}}{2}=-1-\sqrt{3}$ and $w_2=\frac{-2+\sqrt{4+8}}{2}=-1+\sqrt{3}$.  
Sketch of the Graph