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Thread starter 11 months ago
#1
Q1. I was just wondering that since and are just reflections of each other in the x-axis, then can I make the statement:

Since and 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 is greater than 0 for all x in the domain, then by the statement, 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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11 months ago
#2
(Original post by Takeover Season)
Q1. I was just wondering that since and are just reflections of each other in the x-axis, then can I make the statement:

Since and 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 is greater than 0 for all x in the domain, then by the statement, 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 then both functions have the same range.

But if say , then but . These ranges are not the same just because their domain is the same!

A more accurate statement would be that and have the same range provided their domains are reflections about .

I.e. if then . If I take the reflection about the the domain is then and hence 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 instead.
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Thread starter 11 months ago
#3
(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 then both functions have the same range.

But if say , then but . These ranges are not the same just because their domain is the same!

A more accurate statement would be that and have the same range provided their domains are reflections about .

I.e. if then . If I take the reflection about the the domain is then and hence 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 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 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 for all x in the domain. This next part doesn't apply to , 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?
Last edited by Takeover Season; 11 months ago
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11 months ago
#4
(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.
and are actually reflections in the y-axis, but otherwise this is correct.

Sorry, I meant that let's say I have a function 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 for all x in the domain. This next part doesn't apply to , 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 is increasing or not.

Clearly, if then the gradient is increasing (in the direction of x increasing, hence decreasing in the direction of x decreasing).

And if then the gradient is decreasing (in the dir. of x increasing, hence increasing in the direction of x decreasing).

Therefore, in the domain where we have , 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 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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Thread starter 11 months ago
#5
(Original post by RDKGames)
and are actually reflections in the y-axis, but otherwise this is correct.

You're really just asking questions about whether is increasing or not.

Clearly, if then the gradient is increasing (in the direction of x increasing, hence decreasing in the direction of x decreasing).

And if then the gradient is decreasing (in the dir. of x increasing, hence increasing in the direction of x decreasing).

Therefore, in the domain where we have , then the gradient is indeed the gradient is increasing as x decreases, and vice versa.

Yes. Try to think of this in terms of 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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