Let f(x) = x e-x. The maximum value of the function in the interval (0, \[\infty \]) is A. e-1 B. e C. 1 – e-1 D. 1 + e-1

The maximum value of the function $f(x) = x e^{-x}$ in the interval $(0, \infty)$ is $e$.

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 $x = 0$.

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 $x = 0$.

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 $e$.

Here is a graph of $f(x)$:

[asy]
unitsize(1 cm);

draw((0,0)–(10,0));
draw((0,0)–(0,1));

real ticklen = 1.2;
real tickspace = 1.2;
real axisarrowsize = 0.14inch;
real tickdown = -0.1;
real tickdownlength = 0.12inch;
real tickdownbase = -0.12inch;
real wholetickdown = tickdown;
real wholetickdownlength = tickdownlength;
real wholetickdownbase = tickdownbase;
real axisdown = -1.2;
real axisdownlength = 0.2inch;
real axisdownbase = -1.2;
real wholeaxisdown = axisdown;
real wholeaxisdownlength = axisdownlength;
real axisup = 1.2;
real axisuplength = 0.2inch;
real axisupbase = 1.2;
real wholeaxisup = axisup;
real wholeaxisuplength = axisuplength;

real ticklength = 0.08inch;
real tickdownlength = 0.08inch;
real wholetickdownlength = tickdownlength;
real wholetickdownbase = tickdownbase;
real wholeaxisdown = axisdown;
real wholeaxisdownlength = axisdownlength;
real wholeaxisup = axisup;
real wholeaxisuplength = axisuplength;

label(“$x$”, (10,0), E);
label(“$y$”, (0,1), N);

real i;
for (i=-10; i<=10; ++i) {
if (i > -2 && i < 2) {
dot((i,0));
}
draw((i,-0.01)–(i,0.01));
}

real j;
for (j=-1; j<=1; ++j) {
if (abs(j) > 0.1) {
label(“$”+$j, (0,j), S);
}
draw((0,-0.01)–(0,0.01));
}

draw((0,0)–(1,1));
draw((0,0)–(0,e));

draw((0,0)–(exp(-1),exp(-1)));

label(“$e$”, (exp(-1),exp(-1)), S);
[/asy]

As you can see, the maximum value of $f$ occurs at $x = 0$, where $f(0) = e$.