Question

Let $x_0$ be the real number such that $e^{x_0}+x_0=0$. For a given real number $\alpha$, define $g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x+1)}$ for all real numbers $x$.

Then which one of the following statements is TRUE?

Answer 1

g(x_0)=x_0

Answer 2

The given function is $g(x)=\frac{3xe^x+3x-\alpha e^x-\alpha x}{3(e^x+1)}$.

We can factorize the numerator: $g(x)=\frac{3x(e^x+1)-\alpha(e^x+x)}{3(e^x+1)}$ We can split the fraction: $g(x)=\frac{3x(e^x+1)}{3(e^x+1)} - \frac{\alpha(e^x+x)}{3(e^x+1)}$ For $e^x+1 \neq 0$ (which is always true for real $x$ since $e^x > 0$), we can simplify the first term: $g(x)=x - \frac{\alpha(e^x+x)}{3(e^x+1)}$ We are given that $x_0$ is a real number such that $e^{x_0}+x_0=0$. This means $e^{x_0}=-x_0$.

Let's evaluate $g(x)$ at $x=x_0$: $g(x_0)=x_0 - \frac{\alpha(e^{x_0}+x_0)}{3(e^{x_0}+1)}$ Since $e^{x_0}+x_0=0$, the numerator of the second term is zero: $g(x_0)=x_0 - \frac{\alpha(0)}{3(e^{x_0}+1)}$ $g(x_0)=x_0 - 0$ $g(x_0)=x_0$ This result holds true for any real number $\alpha$.

Let's verify this by substituting $e^{x_0}=-x_0$ into the original expression for $g(x_0)$: $g(x_0)=\frac{3x_0e^{x_0}+3x_0-\alpha e^{x_0}-\alpha x_0}{3(e^{x_0}+1)}$ Substitute $e^{x_0}=-x_0$: $g(x_0)=\frac{3x_0(-x_0)+3x_0-\alpha (-x_0)-\alpha x_0}{3(-x_0+1)}$ $g(x_0)=\frac{-3x_0^2+3x_0+\alpha x_0-\alpha x_0}{3(1-x_0)}$ $g(x_0)=\frac{-3x_0^2+3x_0}{3(1-x_0)}$ Factor out $3x_0$ from the numerator: $g(x_0)=\frac{3x_0(1-x_0)}{3(1-x_0)}$ To check if $1-x_0 \neq 0$, consider the function $f(x)=e^x+x$. $f(0)=1$, $f(-1)=1/e-1<0$. Since $f(x)$ is continuous and increasing ($f'(x)=e^x+1>0$), there is a unique root $x_0$ in $(-1, 0)$. Since $x_0 \in (-1, 0)$, $1-x_0$ is in $(1, 2)$ and is not zero. We can cancel the term $(1-x_0)$: $g(x_0)=x_0$ The statement $g(x_0)=x_0$ is true for any given real number $\alpha$.

Without the options provided, we deduce the most likely statement to be tested, based on the problem structure and the universal nature of the derived property, is $g(x_0)=x_0$.