[amp_mcq option1=”e-1″ option2=”e” option3=”1 – e-1″ option4=”1 + e-1″ correct=”option1″]<\/p>\n
The maximum value of the function $f(x) = x e^{-x}$ in the interval $(0, \\infty)$ is $e$.<\/p>\n
To see this, we can first find the derivative of $f$, which is $f'(x) = e^{-x} (1 + x)$. We can then set $f'(x) = 0$ and solve for $x$ to find the critical point of $f$. The critical point is $x = 0$.<\/p>\n
To analyze the critical point, we can use the sign of $f’$ on either side of $x = 0$. We know that $f'(x) > 0$ for all $x > 0$. This means that $f$ is increasing for all $x > 0$. Therefore, the maximum value of $f$ must occur at $x = 0$.<\/p>\n
To find the maximum value of $f$, we can evaluate $f$ at $x = 0$. We find that $f(0) = e$. Therefore, the maximum value of $f$ in the interval $(0, \\infty)$ is $e$.<\/p>\n
Here is a graph of $f(x)$:<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
draw((0,0)–(10,0));
\ndraw((0,0)–(0,1));<\/p>\n
real ticklen = 1.2;
\nreal tickspace = 1.2;
\nreal axisarrowsize = 0.14inch;
\nreal tickdown = -0.1;
\nreal tickdownlength = 0.12inch;
\nreal tickdownbase = -0.12inch;
\nreal wholetickdown = tickdown;
\nreal wholetickdownlength = tickdownlength;
\nreal wholetickdownbase = tickdownbase;
\nreal axisdown = -1.2;
\nreal axisdownlength = 0.2inch;
\nreal axisdownbase = -1.2;
\nreal wholeaxisdown = axisdown;
\nreal wholeaxisdownlength = axisdownlength;
\nreal axisup = 1.2;
\nreal axisuplength = 0.2inch;
\nreal axisupbase = 1.2;
\nreal wholeaxisup = axisup;
\nreal wholeaxisuplength = axisuplength;<\/p>\n
real ticklength = 0.08inch;
\nreal tickdownlength = 0.08inch;
\nreal wholetickdownlength = tickdownlength;
\nreal wholetickdownbase = tickdownbase;
\nreal wholeaxisdown = axisdown;
\nreal wholeaxisdownlength = axisdownlength;
\nreal wholeaxisup = axisup;
\nreal wholeaxisuplength = axisuplength;<\/p>\n
label(“$x$”, (10,0), E);
\nlabel(“$y$”, (0,1), N);<\/p>\n
real i;
\nfor (i=-10; i<=10; ++i) {
\n if (i > -2 && i < 2) {
\n dot((i,0));
\n }
\n draw((i,-0.01)–(i,0.01));
\n}<\/p>\n
real j;
\nfor (j=-1; j<=1; ++j) {
\n if (abs(j) > 0.1) {
\n label(“$”+$j, (0,j), S);
\n }
\n draw((0,-0.01)–(0,0.01));
\n}<\/p>\n
draw((0,0)–(1,1));
\ndraw((0,0)–(0,e));<\/p>\n
draw((0,0)–(exp(-1),exp(-1)));<\/p>\n
label(“$e$”, (exp(-1),exp(-1)), S);
\n[\/asy]<\/p>\n
As you can see, the maximum value of $f$ occurs at $x = 0$, where $f(0) = e$.<\/p>\n","protected":false},"excerpt":{"rendered":"
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