Question

Q1. I was just wondering that since $e^x$ and $e^{-x}$ are just reflections of each other in the x-axis, then can I make the statement:

Since $e^x$ and $e^{-x}$ are just reflection of each other in the x-axis, their range will be the same provided their domain is the same.

So, e.g. if we take the domain as the real numbers, since $e^x$ is greater than 0 for all x in the domain, then by the statement, $e^{-x}$ must also be greater than 0 for all x in the domain.

If I'm to get my toothpick out and examine your statement with precise accuracy, then you are not technically correct.

True, if $x \in \mathbb{R}$ then both functions have the same range.

But if say $x \in [0,1]$, then $e^x \in [1,e]$ but $e^{-x} \in [e^{-1},1]$. These ranges are not the same just because their domain is the same!

A more accurate statement would be that $e^{x}$ and $e^{-x}$ have the same range provided their domains are reflections about $x=0$.

I.e. if $x \in [0,1]$ then $e^{x} \in [0,e]$. If I take the reflection about the $x=0$ the domain is then $y \in [-1,0]$ and hence $e^{-y} \in [0,e]$ as well.



I don't really follow you here. At first you say f(x) is such that f'(x) > 0 but then you go on to suppose the gradient is over $\mathbb{R}$ instead.

Thank you for the explanation, that was great! So, can I say that in general though? e.g.
f(x) and f(-x) are reflections of each other in the x-axis and therefore, provided the domains on which f(x) and f(-x) are defined are reflections about x=0, then both functions f(x) and f(-x) will have the same range.

Sorry, I meant that let's say I have a function $f(x) = e^x$ and the domain is all of the real numbers. Then, it is an increasing function since its gradient is positive over the entire domain as $e^x > 0$ for all x in the domain. This next part doesn't apply to $f(x) = e^x$, but just in general, let's say f(x) was an increasing function throughout its domain of the real numbers.
Then, if I claim its second derivative f''(x) is negative for x < 0 and positive for x > 0. Then, does this mean that in the interval of the domain (- inf, 0), when x decreases, its gradient increases and when x increases, its gradient decreases?

In general, I mean that let's say f''(x) is negative in the interval [a,b]. Then, if x = a and it increases to e.g. x = c which is between x = a and x = b, then I expect the gradient to lower at x = c than it was at x = a?
i.e. as x increases along the interval, the gradient will fall? ... and vice versa... if x decreases from x = b to x = a, the gradient will increase?

$f(x)$ and $f(-x)$ are actually reflections in the y-axis, but otherwise this is correct.




You're really just asking questions about whether $F(x) = f'(x) > 0$ is increasing or not.

Clearly, if $F'(x) > 0$ then the gradient is increasing (in the direction of x increasing, hence decreasing in the direction of x decreasing).

And if $F'(x) < 0$ then the gradient is decreasing (in the dir. of x increasing, hence increasing in the direction of x decreasing).

Therefore, in the domain $x<0$ where we have $F'(x) < 0$, then the gradient is indeed the gradient is increasing as x decreases, and vice versa.




Yes. Try to think of this in terms of $F'(x) = f''(x)$ above as it simplifies the notation a bit and then you don't confuse second derivatives with the notion of increasing/decreasing for the actual function.