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Author Topic: Oliverk94's 3 and 4 Questions  (Read 1279 times)  Share 

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oliverk94

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Oliverk94's 3 and 4 Questions
« on: February 11, 2012, 03:29:31 pm »
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Hey guys.. I'll be posting up questions which I am finding difficulity with and need help in.


1) How do you find the domain and range of a sum and product function

Composite function q:

2) f(x) = 1/(x+a)2 and g(x)=√x, determine the values of a such that f(g(x)) exists.

and

3) If f:R->R, where f(x) = 3- √x and g:R->R, where g(x)=x2-1, show that f o g is not defined. By restricting the domain of g, find a function h such that f o h is defined.


4) How do you do regression? (changing data into linear form) when doing modelling

b^3

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Re: Oliverk94's 3 and 4 Questions
« Reply #1 on: February 11, 2012, 03:53:31 pm »
+1
1) The domain of a sum or product function will be the values of x for which both functions are defined, that is the intersection of the domains of the orginal functions. From the domain you work out the range.

2) R2D1
For the fog(x) to exsit the range of the second must be equal to or a subset of the domain of the first.
In this case





Draw them out if you are having trouble.
Now we need
Currently

Now is , now we need a>0, otherwise Ran g won't be equal to or contained in the domain of f.
https://www.desmos.com/calculator/nhsbi8ijnx
Here the value of -a is what is not included in the domain of g as that is the asymptote (the red line, it should be dotted though)

Hope thats right for those 2 :)
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Somye

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Re: Oliverk94's 3 and 4 Questions
« Reply #2 on: February 12, 2012, 11:03:38 am »
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3) In this case, you're looking for when the range g is a subset or equal to the domain of f
currently,
Range of g = [1,infinity)
Domain of f = (neg. infinity,0]

therefore, range of g is not a subset or equal to the domain of f and so f o g is undefined

so therefore we need to find where the range of g < (or equal) to 0
through a quick sketch we can see that this occurs from [-1,1]
and therefore function h is defined as:[-1,1] -> R where g(x) = x^2 -1

hopefully that makes sense!
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oliverk94

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Re: Oliverk94's 3 and 4 Questions
« Reply #3 on: February 16, 2012, 10:41:37 pm »
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thanks guys.

I need help on finding the domain and range quickly from just the equation.. I can find it very easily when I have the graph but I want to be able to find it from the function/equation itself.. is that possible?

So for example:

y= 3-|x^2+3|

y= -3/(x+4)^2 + 3

y= 3/(x-4) + 3

or any other functions

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Re: Oliverk94's 3 and 4 Questions
« Reply #4 on: February 16, 2012, 11:07:34 pm »
+4
It's always best to visualise the graph (draw it or in your head), as if you have it drawn correctly, you can't really go wrong, where as getting it from the equation there are heaps of ways you can go wrong.

But, most of the time you can find the maximum domains for functions from the function. Have a look at the first couple of rules of pi's guide [GUIDE] Techniques for Sketching Nice-Looking Graphs.

In short, if its a square root, everything under it has to be equal to or greater than 0 (as you can't root a negative number in methods).
If its a 1/something, then that something can't equal 0 (as it would be undefined).
If its log(something) then that something has to be greater than 0 (as you can't log a negative number or a 0).

With Modulus functions from expirence its best to always draw those.

You can also get used to knowing the basic shape of the graph, i.e. for 1/x^2, know that the domain is R\{0} and the range will be R+, and then as that is transformed to y= -3/(x+4)^2 + 3, then transform the domain and range to match, i.e. Domain=R\{-4}, range=(-infinty,3) (careful of that reflection there).

For sine and cos graphs, its always good to start at the vertical mid-point, i.e. vertical shift and find the max and min values for the range from the that vertical shift+-amplitude.

With parabolas, use the turning point form to find the max (if the coefficient of x^2 is negative) or the min (if the coefficient of x^2 is positive). i.e. for y=(x+2)^2  +5, the t.p is a minimum at (-2,5), so the range will be [5,infinity)

Also note that if you are given a restricted domain make sure to check the endpoints as they may be the extreme values for the range.

Most of the time you will find the domain first, then using that domain find the range of the function.

EDIT: Touched a couple of things up :)

Moderator action: removed real name, sorry for the inconvenience
« Last Edit: January 03, 2017, 12:14:49 am by pi »
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oliverk94

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Re: Oliverk94's 3 and 4 Questions
« Reply #5 on: February 16, 2012, 11:10:31 pm »
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Thanks, b^3!

How is maximum domain different from normal domain? I sort of know what it is but can't really explain the difference between the two

b^3

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Re: Oliverk94's 3 and 4 Questions
« Reply #6 on: February 16, 2012, 11:14:52 pm »
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Thanks, b^3!

How is maximum domain different from normal domain? I sort of know what it is but can't really explain the difference between the two
NOTE: I added one or two more things above.

Now the maximal domain is the largest set of values of x for which the function is defined(largest subset of R for which the function is defined). i.e. for y=x Dom=R and for y=1/x, Dom=R\{0}
The "normal" domain is the set of values of x you have for your function. I.e. in real world application problems, you can't have negative distances, so r>0 e.t.c
This "normal" domain may be restricted by the question/circumstances.

Also I probably wouldn't call it "normal" domain, rather just the domain for the function.

EDIT: Thats probably an informal definition for that maybe someone can give a more formal definition for it, *bcub3d ques TrueTears :P
« Last Edit: February 16, 2012, 11:34:14 pm by b^3 »
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oliverk94

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Re: Oliverk94's 3 and 4 Questions
« Reply #7 on: February 16, 2012, 11:28:32 pm »
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thanks