The problem states that given the function $g(x) = \frac{xe^x}{e^x + 1}$, we need to show that $g(a) = a + 1$ and $-0.28 < g(a) < -0.27$. Here, $a$ is the root of the function $f(x)$ in question 1.

1. Problem Description

The problem states that given the function $g(x) = \frac{xe^x}{e^x + 1}$, we need to show that $g(a) = a + 1$ and $-0.28 < g(a) < -0.27$. Here, $a$ is the root of the function $f(x)$ in question

1.

2. Solution Steps

First, we know that $f(a) = 0$. We also know that $f(x) = e^x + x + 1$ from the first problem.

Therefore, $f(a) = e^a + a + 1 = 0$.

This implies $e^a = -(a+1)$.

Now consider the function $g(x) = \frac{xe^x}{e^x+1}$. We need to find $g(a)$.

$g(a) = \frac{ae^a}{e^a+1}$.

Substitute $e^a = -(a+1)$ into the equation for $g(a)$:

$g(a) = \frac{a(-(a+1))}{-(a+1)+1} = \frac{-a(a+1)}{-a} = \frac{-a^2-a}{-a}$.

Since $a \neq 0$ (otherwise $e^0+0+1=0$ which simplifies to $2=0$, which is not true), we can divide by $-a$:

$g(a) = \frac{-a(a+1)}{-a} = a+1$.

The problem also states that $-1.28 < a < -1.27$. Let's use these bounds to determine the bounds for $g(a) = a+1$.

If $a > -1.28$, then $a+1 > -1.28 + 1 = -0.28$.

If $a < -1.27$, then $a+1 < -1.27 + 1 = -0.27$.

Therefore, $-0.28 < a+1 < -0.27$, which implies $-0.28 < g(a) < -0.27$.

3. Final Answer

$g(a) = a+1$ and $-0.28 < g(a) < -0.27$.