[amp_mcq option1=”2″ option2=”1″ option3=”0″ option4=”-1″ correct=”option1″]<\/p>\n
The maximum occurs when $x=0$.<\/p>\n
To find the maximum, we can take the derivative of $f(x)$ and set it equal to zero. The derivative of $f(x)$ is $f'(x) = 2xe^{-x} – x^2e^{-x} = x(2e^{-x} – x)$. Setting $f'(x)=0$, we get $x(2e^{-x} – x) = 0$. Solving for $x$, we get $x=0$ or $x=2e^{-x}$.<\/p>\n
Since $e^{-x} > 0$ for all real numbers, $2e^{-x} > x$ for all real numbers except $x=0$. Therefore, the only solution to $x(2e^{-x} – x) = 0$ is $x=0$.<\/p>\n
We can also find the maximum by looking at the graph of $f(x)$. The graph of $f(x)$ is shown below.<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
real ticklen=3;
\nreal tickspace=2;
\nreal axisarrowsize=0.14inch;
\nreal tickdown=-0.14inch;
\nreal wholetickdown=-0.2inch;
\nreal wholetickdownlength=0.14inch;
\nreal wholetickdownbase=-0.24inch;
\nreal wholetickdownextend=0.14inch;
\nreal wholetickdownextendbase=-0.34inch;
\nreal tickdownbase=-0.14inch;
\nreal tickdownextend=0.14inch;
\nreal tickdownextendbase=-0.24inch;
\nreal ticklength=0.14inch;
\nreal axispen=black+0.7bp;
\nreal tickpen=black+0.7bp;
\nreal xmin=-2;
\nreal xmax=2;
\nreal ymin=-2;
\nreal ymax=2;
\nreal xstep=0.2;
\nreal ystep=0.2;<\/p>\n
draw((xmin,0)–(xmax,0),axispen);
\ndraw((0,ymin)–(0,ymax),axispen);<\/p>\n
label(“$x$”,(xmax,-0.1),E);
\nlabel(“$y$”,(-0.1,ymax),N);<\/p>\n
for (int i=xmin+xstep; i<xmax; i+=xstep) {
\n draw((i,0.01)–(i,-0.01));
\n}<\/p>\n
for (int i=ymin+ystep; i<ymax; i+=ystep) {
\n draw((0.01,i)–(-0.01,i));
\n}<\/p>\n
real g(real x) {
\n return x^2*exp(-x);
\n}<\/p>\n
draw(graph(g,xmin,xmax),red);<\/p>\n
draw((0,g(0))–(0.1,g(0)),dashed);<\/p>\n
label(“$f(x)$”,(1.2,1.2),S);
\nlabel(“$y=x^2e^{-x}$”,(0.5,1.2),S);
\nlabel(“$x=0$”,(0,g(0)),S);
\n[\/asy]<\/p>\n
As you can see, the graph of $f(x)$ reaches its maximum at $x=0$.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”2″ option2=”1″ option3=”0″ option4=”-1″ correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20197","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n