Let's solve the problem step by step:

Problem:

Given $f(x) = 3 \cdot 5^{(2(x-1)^2 - 18)} - 75$, we address the following:

(a) Hallar el $\text{Dom}(f)$:

The domain of $f(x)$ is determined by analyzing if there are any restrictions on $x$ for which $f(x)$ is undefined. In this case, $5^{(2(x-1)^2 - 18)}$ is always defined for all real $x$ because $5^z$ is defined for any real $z$. Thus, there are no restrictions.

Answer: $\text{Dom}(f) = \mathbb{R}$

(b) Mostrar que $f : \text{Dom}(f) \to \mathbb{R}$ no es inyectiva:

A function is injective if every $f(a) = f(b)$ implies $a = b$. Here, note that the term $(x-1)^2$ is symmetric about $x = 1$, so $f(x)$ will take the same value for $x = a$ and $x = 2 - a$. Thus, $f(x)$ is not injective over $\mathbb{R}$.

Proof: Let $x_1 = 0$ and $x_2 = 2$: $f(0) = 3 \cdot 5^{2(0-1)^2 - 18} - 75 = 3 \cdot 5^{-16} - 75$ $f(2) = 3 \cdot 5^{2(2-1)^2 - 18} - 75 = 3 \cdot 5^{-16} - 75$ Since $f(0) = f(2)$ but $0 \neq 2$, $f(x)$ is not injective.

Conclusion: $f(x)$ is not injective on $\mathbb{R}$.

(c) Probar que $f : (-\infty, 1] \to \mathbb{R}$ es inyectiva:

On the interval $(-\infty, 1]$, $(x-1)^2$ is strictly decreasing because the square of a number gets smaller as $x \to 1^-$. Thus, $2(x-1)^2 - 18$ is strictly decreasing, and so is $5^{2(x-1)^2 - 18}$. The subtraction of 75 and multiplication by 3 do not change this monotonic behavior.

Proof: Let $x_1, x_2 \in (-\infty, 1]$ with $x_1 < x_2$. Then: $(x_1 - 1)^2 > (x_2 - 1)^2 \implies 2(x_1-1)^2 - 18 > 2(x_2-1)^2 - 18$ $\implies 5^{2(x_1-1)^2 - 18} > 5^{2(x_2-1)^2 - 18} \implies f(x_1) > f(x_2)$ Thus, $f(x)$ is strictly decreasing on $(-\infty, 1]$ and is injective.

Conclusion: $f(x)$ is injective on $(-\infty, 1]$.

(d) **Hallar la expresión de ( f^{-1} : \text{Im}(f) \