As x increased from \[ – \infty \] to \[\infty \] , the function \[{\rm{f}}\left( {\rm{x}} \right) = \frac{{{{\rm{e}}^{\rm{x}}}}}{{1 + {{\rm{e}}^{\rm{x}}}}}\] A. monotonically increases B. monotonically decreases C. increases to a maximum value and then decreases D. decreases to a minimum value and then increases

The correct answer is: A. monotonically increases.

The function $f(x) = \frac{e^x}{1 + e^x}$ is monotonically increasing because its derivative $f'(x) = e^x \left( 1 – \frac{1}{1 + e^x} \right) > 0$ for all $x$.

To see this, we can use the fact that $e^x > 0$ for all $x$. Then, we can simplify $f'(x)$ as follows:

$$f'(x) = e^x \left( 1 – \frac{1}{1 + e^x} \right) = e^x \left( \frac{1 + e^x – 1}{1 + e^x} \right) = e^x \left( \frac{e^x}{1 + e^x} \right) = e^x.$$

Since $e^x > 0$ for all $x$, we know that $f'(x) > 0$ for all $x$. This means that $f(x)$ is increasing for all $x$.

In other words, as $x$ increases from $-\infty$ to $\infty$, the value of $f(x)$ also increases.