I hope you are all doing well. I am currently struggling with a problem and I would greatly appreciate your help in understanding the concept and learning how to solve similar problems.
The problem is from a TEST BAC in Algeria and the question is as follows:
Prove that the function $C_f$ has a tangent line that is parallel to the asymptote $y = x$ at a certain $x_0$ value, which we need to find.
The given information is as follows:
- $f(\alpha) = \alpha + 1$
- $g(\alpha) = 0$
- $f(x) = \frac{xe^x}{e^x + 1}$
- $f'(x) = \frac{e^x \cdot g(x)}{(e^x + 1)^2}$
- $g(x) = x + 1 + e^x$
$C_f$ represents the graph of function $f(x)$ and $\alpha$ is a solution to $g(x)$
I would like to understand the steps involved in proving the existence of a tangent line for the function $C_f$ that is parallel to the asymptote $y = x$ at a certain $x_0$ value. Additionally, I need guidance on how to determine the value of $x_0$ based on the given information.
Thank you for your assistance in solving this problem and providing insights for similar problems.
You are asking whether $f$ has a tangent line that is parallel to $y=x$, i.e. has slope $1$. Then we need to determine if there exists ${x_0}$ such that $f'({x_0})=1$, i.e. $$ e^{x_0}({x_0}+1+e^{x_0})=(e^{x_0}+1)^2. \tag{1} $$ Dividing by $e^{x_0}\neq 0$ and simplifying, we have $$ -{x_0}+e^{-{x_0}}+1=0. \tag{2} $$ (The converse is also true.) This is reminiscent of the given $g(\alpha)=0$, i.e. $$ \alpha+e^\alpha+1=0. $$ If ${x_0}=-\alpha$, then we have $(2)$ and hence $(1)$. This shows that $f$ indeed has a tagent line with slope $1$ at the point $x_0=-\alpha$.