Is one of the reasons we know $ y = e^{2x}$ is always positive that it is a square? One of my Calculus assignments asks how we can tell that $ y = e^{2x}$ is always positive.
The options given are:


*

*It's an exponential function with a base $> 0$.

*It's a square.

*Its derivative is always positive.


I said the only reason was 1. But I was told that number 2 is also a valid reason given that $y = e^{2x} = (e^x)^2$ is a square.
If $e$ were negative, however, wouldn't the function be negative at $x = 1/2$? 
Is 2 a valid reason to conclude that $e^{2x}$ is always positive?
 A: The reason is (1): any positive real number to any power is positive. Why? Definition and/or reason to define logarithms and stuff (I can't say what you know and what you don't...)
A: You should be careful to note if your professor means "positive" or "nonnegative". A square (of a real number) cannot be negative, but it can be zero.
A: The function $s^x$ is not defined (among reals) unless $s\ge 0$. 
For an $s<0$, the term $(s^{1/2})^2$ is simply not defined (among reals).
A: Indeed $e^{2x} = (e^x)^2$ is positive because it is a square of $e^x$ which is positive.
Or $e^{2x} = (e^2)^x$ is a positive function because the base is positive.
I might go with (1) since with (2) you have to also note that $e^x$ is positive.
As a sidenote and since it is sort of related: You have to be careful with negative base numbers. With positive base numbers the exponential is always defined and you get a real number. We can of course make sense of $(-1)^7$ which is negative or even $(-1)^{\frac{1}{2}} = i$, but we don't necessarily get a real number and so saying that it is positive doesn't even make sense. Also, how would you in general define a function $s^{x}$ for any negative $s$? What is for example $(-1)^\pi$ ?
The usual rules of exponents even get messed up. See for example
$$
1 = (1^2)^{\frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = ((-1)^{\frac{1}{2}})^2 = i^2 = -1.
$$
What went wrong? See this Wikipedia article for a bit more about exponentiation.
A: Hint
If the option 2. is true what can we say about
$$(-1)^{1}=(-1)^{2\times\frac{1}{2}}=i^2?$$
A: Point 2 is valid by itself for the purposes of providing a way to show that $e^{2x}$ (and, indeed, $e^x$) are nonnegative.
The fact than an exponential is nonnegative everywhere is inextricably linked to the fact that when we double its argument, its value squares!
However, point 2 contains an embedded assumption: that $e^x$, for real $x$ is a real value (either positive or negative, but not complex). It should be stated.
Suppose that some real-valued function $f$ has the property that $f(2x) = {f(x)}^2$, over all real $x$. Can $f(x)$ be negative? Clearly not, and it's easy to see if we rewrite the relation as $f(x) = {f(x/2)}^2$.  For all $x$, $f(x)$ is the square of some real number, and so it must be nonnnegative.
The function $e^x$ is readily identified as being such an $f$.
Furthermore, if we have a function $g$ such that $g(x) = f(2x)$, then $g$ is just $f$ scaled down by a factor of two, horizontally. If $f$ is nonnegative (or positive), then $g$ is likewise, and vice versa, since horizontal scaling has no effect on that.
Point 2, however, requires additional arguments to that $e^x$ is never zero, and thus positive.
A: Assuming a number $x$ is positive if $sgn(x) \ge 0$ where $sgn(x)$ is the signum function 
For Complex number's an equivalent definition would be
$$
\operatorname{csgn}(z)= \begin{cases} 1 & \text{if } \Re(z) > 0, \\ -1 & \text{if } \Re(z) < 0, \\ sgn(\Im(z)) & \text{if } \Re(z) = 0 \end{cases} 
$$
Consider the expression
$$y = e^{2x}$$
Let $x=0+i$, so we have
$$\operatorname{csgn}(y) = \operatorname{csgn}(e^{2x})=\operatorname{csgn}(e^{2i})=-1$$
A: No, 2) is not, because being a square does not exclude the possibility it is zero somewhere. But it is enough to guarantee it nonnegative.
