Proof that 2 funtion of the type $f_t(x) = \frac{1}{t} \cdot e^{-tx²} $ don't intersect so as the titles states I wan't to proof that no two functions $f_t$ of type $$f_t(x) = \frac{1}{t} \cdot e^{-tx²} \; \text{given that} \; t>0$$ share a point.
This is a question from my textbook.
However I come to the conclusion that there should be 2 functions should share some point.
$$f_a(x) = \frac{1}{a} \cdot e^{-ax²}$$
$$f_b(x) = \frac{1}{b} \cdot e^{-bx²}$$
Let $f_a(x) = f_b(x)$
$$\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²}$$
$$\frac{1}{a} \cdot e^{a} = \frac{1}{b} \cdot e^{b}$$
$$ae^{-a}=be^{b}$$
In order to proof that the 2 functions don't share a point I would need to proof that a=b
This would mean that the function $g(x)=xe^x$ is a one to one function. However this is not true!
Now, all continuous one-one functions have to be monotonic (strictly increasing or decreasing). But we know that this is not the case for $g(x)$ since $\lim_{x \to 0} g(x)=\lim_{x \to +\infty}f(x)=0$ whereas $f(1)>0$
Where is my mistake?
 A: Let suppose that for some $c\neq 0$ we have
$$\frac{1}{a} \cdot e^{-ac^2} = \frac{1}{b} \cdot e^{-bc^2}\iff \frac{\log b-\log a}{a-b}=c^2<0\ \text{contradiction}$$
A: 
$\dfrac{1}{a} \cdot e^{-ax²} = \dfrac{1}{b} \cdot e^{-bx²} \implies \dfrac{1}{a} \cdot e^{a} = \dfrac{1}{b} \cdot e^{b}.$

This step is not correct. It should be
$$\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²} \implies 
\frac{b}{a} = e^{(a-b)x^2}.$$
Taking natural log on both sides given $a,b>0$:
$$
(a-b)x^2 = \ln b-\ln a.
$$
Without loss of generality assuming $a>b>0$, then
$$
(a-b)x^2 \geq 0, \;\text{ and }\; \ln b-\ln a < 0.
$$
Contradiction.
A: The question has been amply answered, so we give another proof of non-intersection.
Let $x$ be fixed.  Differentiate with respect to $t$. The derivative is negative. So our function, for fixed $x$, is a decreasing function of $t$. 
A: I don't understand how you go from the $\frac{1}{a} \cdot e^{-ax²} = \frac{1}{b} \cdot e^{-bx²}$ step to the next.
Anyway, here is a hint:
\begin{align}
\frac{1}{a} \cdot e^{-ax²} &= \frac{1}{b} \cdot e^{-bx²}\\
\frac{b}{a} &=  e^{ax^2-bx^2}\\
\text{If } b>a\\
\underbrace{\frac{b}{a}}_{>1}&=  e^{\underbrace{(a-b)}_{<0}x^2}\\
\end{align}
Contradiction! (The exponent of a negative number can never exceed 1)
Can you similarly prove a contradiction for $b<a$
