point belong to all tangents of $xe^x$.

Question

Given a function $f$ defined on $\mathbb{R}$ as : $f(x)=xe^{-x}$, it graph is $(C)$.

Question : are there any point belong to all the tangents of $(C)$ ?

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This is a part from a generalized problem : $f_{\lambda} (x)=(x+\lambda) e^{-x} $ and it graph is $(C_{\lambda})$.

But, is it correct ? I mean, are there any such points ?

Answer 1

There is no point common to all the tangents of $C$, whether on the curve or not. The tangent line at $x=a$ is $y=(e^{-a}-ae^{-a})x+a^2e^{-a}.$ For $a=1$, this is horizontal at $e^{-1}$. For $a=0,$ this is $y=x$. These intersect at $(e^{-1},e^{-1})$ so if there is a common point, this is it. It is not on the tangent at $a=0.5$

Answer 2

For a function $f$ the tangent at $(a, f(a))$ is $\frac{y-f(a)}{x-a} = f'(a)$ or $y = (x-a)f'(a) + f(a) $.

If there is a point that all these tangents pass through, then the set of equations

$\begin{align} y &= (x-a)f'(a) + f(a)\\ y &= (x-b)f'(b) + f(b)\\ \end{align} $

should have the same solution $(x, y)$ for any $a$ and $b$.

Subtracting them, $0 = x(f'(a)-f'(b))-a f'(a) + b f'(b) + f(a)-f(b)$ or $$x = \frac{-a f'(a) + b f'(b) + f(a)-f(b)}{f'(b)-f'(a)}$$ and $y$ is messier.

Requiring $x$ to be the same for all $a$ and $b$ would result in a differential equation involving $f$ that would be gotten by letting $b \to a$. I'm not willing to do this, so I'll work with the supplied function.

For $f(x) = x e^{-x}$, $f'(x) = -x e^{-x} + e^{-x} = (1-x) e^{-x}$, so

$\begin{align} x &= \frac{-a(1-a)e^{-a} + b(1-b)e^{-b} + a e^{-a}-be^{-b}}{(1-b)e^{-b}-(1-a)e^{-a} }\\ &= \frac{(a-a(1-a))e^{-a} + (b(1-b)-b)e^{-b}}{(1-b)e^{-b}-(1-a)e^{-a} }\\ &= \frac{a^2e^{-a} - b^2e^{-b}}{(1-b)e^{-b}-(1-a)e^{-a} }\\ \end{align} $

For $a=0$ this is $\frac{-b^2e^{-b}}{(1-b)e^{-b}-1 }$

Since this is not the same for all $a$ and $b$, there is no such point.