Are these 2 piecewise functions equivalent

Question

$$f(x)= \begin{cases} e^{-(x+2)}&\text{if}\, -2 < x < \inf\\ 0&\text{otherwise} \end{cases}$$ and $$f(x)= \begin{cases} e^{-x}&\text{if }\, 0 < x < \inf\\ 0&\text{otherwise} \end{cases}$$

Are these 2 piecewise functions equivalent?

Answer 1

Remembering function transformations. Suppose:

$$f(x)=e^{-x}\quad \quad f(x+2)=e^{-(x+2)}$$

$f(x+2)$ the parent function $e^{-x}$ shifted to the left $2$ units. So, it is not equal to the other piecewise function: the domains of each piecewise function encompass the same curve, but are not in the same place because the first is transformed. As an example, try plugging in some $x$-values, such as $x=1$:

$$e^{-(1+2)} \neq e^1$$