Yatayyat
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I have been doing a few of these questions and I'm unsure on what the range of what some of these functions could be?

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It also states that I need to give a reasoning for why the original domain is wrong

These are what I have gave so far:

a) x cannot be >0 as this includes x=3, which therefore gives the function an undefined result when x=3 (i.e. there is a vertical asymptote at x=3)
If x>0 is allowed then the function is no longer total hence we must restrict the domain appropriately to ensure function is total

b) x not equal to 0 is false because this lets x to be all negative real numbers. If so then inside the function's numerator square root can yield a negative number which is not possible (result would be undefined.)

c) x not equal to 0 won't work as inside the function's denominator square root, some numbers could yield a negative number. Taking the square root of a negative number is undefined which would cause the entire result to be undefined.

d) x>=0 is is false as this implies that you can take the natural log of 0 which is not true. ln(0) is undefined.

Any help would be great. Thanks
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RDKGames
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(Original post by Yatayyat)
a) x cannot be >0 as this includes x=3, which therefore gives the function an undefined result when x=3 (i.e. there is a vertical asymptote at x=3)
If x>0 is allowed then the function is no longer total hence we must restrict the domain appropriately to ensure function is total
Yep. For the range, it's obvious from the sketch that (0,\infty) is part of it. For the branch going towards -ve infinity, you need to figure out what they coordinate of the top of it is, but again from the diagram, its just where the function intercepts the y axis.

Then the overall range is the union of these answers.

b) x not equal to 0 is false because this lets x to be all negative real numbers. If so then inside the function's numerator square root can yield a negative number which is not possible (result would be undefined.)
Not sure what you mean by the first sentence, but your working out is good.

For the range, observe that it is definitely y \geq 0, but is there an upper bound to this function ?? At x=-9/7 we have y=0, then at x=0 we have y=3/4. Hence the function has increased, but also as x\to \infty we have that y \to 0 therefore the function increases to 3/4 (and perhaps a bit more ??) before it decides to turn around and tend towards y=0 again. As this is a continuous function, there must be a maximum for you to find. Do that and this will be your upper bound.

c) x not equal to 0 won't work as inside the function's denominator square root, some numbers could yield a negative number. Taking the square root of a negative number is undefined which would cause the entire result to be undefined.
For the range here, note that it can be rewritten as \displaystyle \sqrt{1-\dfrac{1}{x-1}}. I'm sure you can tell me what the range is for 1-\dfrac{1}{x-1} so then taking the square root will leave the range as...??

d) x>=0 is is false as this implies that you can take the natural log of 0 which is not true. ln(0) is undefined.
So x \neq 0. Fine. There are two other values for which the function \dfrac{1}{\ln (x^2)} is undefined [HINT: they make the denominator zero.] Hence the domain is...??

For the range, well you just need to observe what sort of values the function takes inbetween the points where its undefined.
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Yatayyat
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(Original post by RDKGames)
Yep. For the range, it's obvious from the sketch that (0,\infty) is part of it. For the branch going towards -ve infinity, you need to figure out what they coordinate of the top of it is, but again from the diagram, its just where the function intercepts the y axis.

Then the overall range is the union of these answers.



Not sure what you mean by the first sentence, but your working out is good.

For the range, observe that it is definitely y \geq 0, but is there an upper bound to this function ?? At x=-9/7 we have y=0, then at x=0 we have y=3/4. Hence the function has increased, but also as x\to \infty we have that y \to 0 therefore the function increases to 3/4 (and perhaps a bit more ??) before it decides to turn around and tend towards y=0 again. As this is a continuous function, there must be a maximum for you to find. Do that and this will be your upper bound.



For the range here, note that it can be rewritten as \displaystyle \sqrt{1-\dfrac{1}{x-1}}. I'm sure you can tell me what the range is for 1-\dfrac{1}{x-1} so then taking the square root will leave the range as...??



So x \neq 0. Fine. There are two other values for which the function \dfrac{1}{\ln (x^2)} is undefined [HINT: they make the denominator zero.] Hence the domain is...??

For the range, well you just need to observe what sort of values the function takes inbetween the points where its undefined.
Ok so I think I have got the range already for part a)
and the domain and range for part d)

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For part b) to work out the upper bound, do you suggest I find the 1st order derivative of the function, then set that equal to zero to find the x coordinate of the upper bound value. Then to get the y coordinate, I just sub in the x coordinate I got earlier into f(x)? I can't see what else I could do apart from that

For part c) I really don't know how you expressed the function to be rewritten as sqrt(1 – 1/ (x–1))
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RDKGames
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(Original post by Yatayyat)
Ok so I think I have got the range already for part a)
But this function doesn't intercept the y-axis at y=0 ??[/QUOTE]

For part b) to work out the upper bound, do you suggest I find the 1st order derivative of the function, then set that equal to zero to find the x coordinate of the upper bound value. Then to get the y coordinate, I just sub in the x coordinate I got earlier into f(x)? I can't see what else I could do apart from that
Exactly that.


For part c) I really don't know how you expressed the function to be rewritten as sqrt(1 – 1/ (x–1))
Sorry I misread the function.

For \displaystyle y=\dfrac{\sqrt{x}}{\sqrt{x^2-1}} you have that x > 1 as the domain, and y is definitely non-negative as it's defined in terms of square roots.

As x\to 1_+, what happens to y ??

As x \to \infty, what happens to y ??

This should trace out the range for you.

part d
Domain not quite right. This function is defined everywhere on \mathbb{R} except at those three points.

For the range, you got it.
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Yatayyat
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(Original post by RDKGames)
But this function doesn't intercept the y-axis at y=0 ??


Exactly that.




Sorry I misread the function.

For \displaystyle y=\dfrac{\sqrt{x}}{\sqrt{x^2-1}} you have that x > 1 as the domain, and y is definitely non-negative as it's defined in terms of square roots.

As x\to 1_+, what happens to y ??

As x \to \infty, what happens to y ??

This should trace out the range for you.



Domain not quite right. This function is defined everywhere on \mathbb{R} except at those three points.

For the range, you got it.[/QUOTE]

Ok so for a) I know my range can be all real numbers except for zero

And for b) I have tried to differentiate it using quotient rule and chain rule but the fraction seems to get really messy and it seems like a lot of work just to get the range

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For c) because there is a vertical asymptote at x=1 I believe, as x approaches as close to 1, y shoots up to +ve infinity and as x tends to infinity then y tends to 0 but never would be 0, so my range would be [0, +ve infinity)?

My domain for d) is (–ve infinity, –1) U (–1,0) U (0,1) U (1, +ve infinity)? I realised I was missing out on one of the values before
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RDKGames
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(Original post by Yatayyat)
Ok so for a) I know my range can be all real numbers except for zero
Okay yep, I was confused before as I thought they wanted the range BEFORE you correct the domain. But yes, under the domain \mathbb{R} \setminus \{ 3 \} the range is indeed \mathbb{R} \setminus \{ 0 \}.

And for b) I have tried to differentiate it using quotient rule and chain rule but the fraction seems to get really messy and it seems like a lot of work just to get the range
For y'=0 we just require that from y = \dfrac{u}{v} we have u'v = uv'.

Hence this becomes

\dfrac{7(x^2+4)}{2\sqrt{7x+9}} = 2x\sqrt{7x+9} \iff 7(x^2 + 4) = 4x(7x+9)

\iff 21x^2 + 36x - 28 = 0

and pick the root that is in the domain. Hence get the y value.

For c) because there is a vertical asymptote at x=1 I believe, as x approaches as close to 1, y shoots up to +ve infinity and as x tends to infinity then y tends to 0 but never would be 0, so my range would be [0, +ve infinity)?
Pretty much, though you've just said "never would be 0" and then go ahead and put a square bracket. Which one is it??

My domain for d) is (–ve infinity, –1) U (–1,0) U (0,1) U (1, +ve infinity)? I realised I was missing out on one of the values before
Yep, or in a more compact form, \mathbb{R} \setminus \{0, \pm 1 \}.
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Yatayyat
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(Original post by RDKGames)
Okay yep, I was confused before as I thought they wanted the range BEFORE you correct the domain. But yes, under the domain \mathbb{R} \setminus \{ 3 \} the range is indeed \mathbb{R} \setminus \{ 0 \}.



For y'=0 we just require that from y = \dfrac{u}{v} we have u'v = uv'.

Hence this becomes

\dfrac{7(x^2+4)}{2\sqrt{7x+9}} = 2x\sqrt{7x+9} \iff 7(x^2 + 4) = 4x(7x+9)

\iff 21x^2 + 36x - 28 = 0

and pick the root that is in the domain. Hence get the y value.



Pretty much, though you've just said "never would be 0" and then go ahead and put a square bracket. Which one is it??



Yep, or in a more compact form, \mathbb{R} \setminus \{0, \pm 1 \}.
For b) I have tried out the algebra too and understand everything, but I can't seem to figure out why u'v = uv' in the first place?
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I get my correct value of x which is in the domain to be 0.581. Giving my range to be [0, 0.581]

And for c), my mistake the range should be (0, +ve infinity) instead?

Edit: Nevermind, it makes sense to me now how you got uv' = vu'

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Last edited by Yatayyat; 1 year ago
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RDKGames
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(Original post by Yatayyat)
For b) I have tried out the algebra too and understand everything, but I can't seem to figure out why u'v = uv' in the first place?
From the quotient rule, if y'=0 then the numerator is =0. Set the numerator =0 then my equality follows.

I get my correct value of x which is in the domain to be 0.581. Giving my range to be [0, 0.581]
You should stick with exact values. Also 0 isnt included in the range.

And for c), my mistake the range should be (0, +ve infinity) instead?
Yes.
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Yatayyat
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(Original post by RDKGames)
From the quotient rule, if y'=0 then the numerator is =0. Set the numerator =0 then my equality follows.



You should stick with exact values. Also 0 isnt included in the range.



Yes.
So range of b) sticking with exact values and excluding 0 gives (0, (–18+4sqrt(57))/21] ?
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RDKGames
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(Original post by Yatayyat)
So range of b) sticking with exact values and excluding 0 gives (0, (–18+4sqrt(57))/21] ?
Yep.
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Yatayyat
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(Original post by RDKGames)
Yep.
Thank you tremendously for your time in helping me!

I was getting pretty stuck on when how I would be able to find the range for b) until you told me that numerator is simply equal of zero which simplified things a lot more
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