Takeover Season
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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.

Q2. If I have a function f(x) and I claim it's gradient is positive throughout its domain i.e. it is an increasing function. Let's say the gradient is all of the real numbers again. Then, if I claim it's second derivative f''(x) is negative when x is negative. Does this mean that in the range (- inf, 0), when x decreases, its gradient increases and when x increases, its gradient decreases?
Last edited by Takeover Season; 11 months ago
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RDKGames
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(Original post by Takeover Season)
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.

Q2.
If I have a function f(x) and I claim it's gradient is positive throughout its domain i.e. it is an increasing function. Let's say the gradient is all of the real numbers again. Then, if I claim it's second derivative f''(x) is negative when x is negative. Does this mean that in the range (- inf, 0), when x decreases, its gradient increases and when x increases, its gradient decreases?
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.
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(Original post by RDKGames)
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.

[b]

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?
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RDKGames
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(Original post by Takeover Season)
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.
f(x) and f(-x) are actually reflections in the y-axis, but otherwise this is correct.


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 &gt; 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?
You're really just asking questions about whether F(x) = f'(x) &gt; 0 is increasing or not.

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

And if F'(x) &lt; 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&lt;0 where we have F'(x) &lt; 0, then the gradient is indeed the gradient is increasing as x decreases, and vice versa.


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?
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.
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(Original post by RDKGames)
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) &gt; 0 is increasing or not.

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

And if F'(x) &lt; 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&lt;0 where we have F'(x) &lt; 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.
Thank you, I am so happy with your explanations - I feel I have a better way of remembering it now. The notation often confuses me and makes me make incorrect statements and therefore draw a graph incorrectly. I appreciate your help a lot.
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