Is $a_2 \left( {-x \exp(-2x) \over 2} - {\exp(-2x) \over 4} + {1 \over 4} \right)$ a one-one function

Question

I have a function $\displaystyle a_2 \left( {-x \exp(-2x) \over 2} - {\exp(-2x) \over 4} + {1 \over 4} \right)$.

I need to check whether it is one-one.
I know that to show a function is one-one, we must show that if $f(x) = f(y)$ then $x=y$.
Therefore, $$a_2 \left( {-x \exp(-2x) \over 2} - {\exp(-2x) \over 4} + {1 \over 4} \right) = a_2 \left( {-y \exp(-2y) \over 2} - {\exp(-2y) \over 4} + {1 \over 4} \right) $$ $$\left( {-x \exp(-2x) \over 2} - {\exp(-2x) \over 4} \right) = \left( {-y \exp(-2y) \over 2} - {\exp(-2y) \over 4} \right) $$ $$\left[ ({-2x \exp(-2x)}) - ({\exp(-2x)}) \right] = \left[ ({-2y \exp(-2y)}) - ({\exp(-2y)}) \right] $$ $$\left[ ({-2x \exp(-2x)}) - ({\exp(-2x)}) \right] = \left[ ({-2y \exp(-2y)}) - ({\exp(-2y)}) \right] $$ $$\left( {x \exp(-2x)} \right) = \left({y \exp(-2y)}) \right) $$

I believe, $x \exp(x)$ is not one-one...I'm I correct?

Answer 1

You have a slight mistake in your calculations. You should have \begin{align*} &-2 xe^{-2x}- e^{-2x} = -2 ye^{-2y}- e^{-2y} \\ \color{blue}{\implies} &-(2x+1) e^{-2x}\cdot \frac{1}{e} = -(2y+1) e^{-2y}\cdot \frac{1}{e} \\ \color{blue}{\implies} &-(2x+1) e^{-(2x+1)} = -(2y+1) e^{-(2y+1)} \end{align*} So although the question does reduce to checking when $xe^{x}$ is invertible, the argument, in this case, is $-(2x+1)$, and not just $-2x$.

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The function $xe^{x}$ is not one-two-one for all real values of $x$. However, it is one-two-one if you restrict the domain.

If you take $-(2x+1),-(2y+1) \ge -1 \color{blue}{\implies} x,y >0$ then the principal branch of the Lambert-W function $W_0$ does give you an inverse for $xe^{x}$, and thus, on the interval $\left[0,\infty\right)$ your function is one-two-one.

Similarly, if your domain is $(-\infty, 0]$ the function $W_{-1}$ gives you an inverse. But if you're taking the entire real number line as your domain, a simple plot of the function combined with the horizontal-line test tells you the function is not one-to-one on this domain.

Answer 2

Your function is continuous. A continuous function is one-to-one if and only if it is strictly decreasing or strictly increasing.

In your case, the function is differentiable so we only need to see its first derivative which by simple computation turns out to be $a_2$ times $xe^{-2x}$. The latter is positive iff $x > 0$ and negative iff $ x < 0$.

Answer 3

It's convenient to consider the core expression $-2x\exp(-2x)-\exp(-2x)+1$, let $u=-2x$, and rewrite this function as

$$f(u)=(u-1)e^u+1$$

We see that $f(0)=0$, $\lim_{u\to+\infty}f(u)=+\infty$, and $\lim_{u\to-\infty}f(u)=1$. Consequently each value between $0$ and $1$ occurs at least twice, once with $u\gt0$ and once with $u\lt0$. So the original function is not one-to-one.