- #1
Air
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I don't understand what the range is about. Can someone please explain? Thanks in advance.
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I don't understand what the range is about?
- Context: High School
- Thread starter Air
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Discussion Overview
The discussion centers around understanding the concept of range in mathematical functions, including methods for calculating it and the implications of domain restrictions. Participants explore various approaches to determine the range, including the use of inverse functions and calculus, while also addressing potential pitfalls in these methods.
Discussion Character
- Exploratory
- Technical explanation
- Mathematical reasoning
- Debate/contested
Main Points Raised
- Some participants explain that the range is the set of values that a function can output, with examples provided to illustrate this concept.
- One participant suggests calculating the range by finding the inverse of the function and determining the domain of that inverse.
- Another participant warns about the complications that arise when dealing with certain functions, particularly regarding the absolute value in inverses.
- There is a discussion about the necessity of considering domain restrictions when calculating ranges, with some participants noting that their methods may not account for these restrictions.
- One participant emphasizes that the function must be injective (one-to-one) for the method of finding the range through the inverse to work correctly.
- Another participant provides a detailed example of calculating the range for a specific function, illustrating the steps taken and the reasoning behind them.
Areas of Agreement / Disagreement
Participants express differing views on the best methods to determine the range, with some advocating for visual inspection while others prefer algebraic approaches. There is no consensus on a single method, and the discussion remains unresolved regarding the most effective techniques.
Contextual Notes
Some participants note limitations in their approaches, such as not accounting for domain restrictions or the injectivity of functions, which may affect the accuracy of their conclusions.
Who May Find This Useful
This discussion may be useful for students and individuals seeking to deepen their understanding of mathematical functions, particularly in relation to the concepts of range and domain, as well as those interested in exploring different methods of analysis in mathematics.
I don't understand what the range is about. Can someone please explain? Thanks in advance.
- #2
prasannapakkiam
The domain is the set values of the function in which the value would yeild a real number. for example: f(x)=5/x, x is not part of the domain as 5/0 is not real.
The same, the range is the set of value from which if you put a number into the function these values can be found. For example: f(x)=x^2+1. The range is: f(x)=>1. This is because you cannot obtain numbers below 1 - put any number into the function and you will not be able to numbers below 0.
to calculate the range, you simply find the inverse of the function and find the domain of the inverse function - or use differential calculus...
The same, the range is the set of value from which if you put a number into the function these values can be found. For example: f(x)=x^2+1. The range is: f(x)=>1. This is because you cannot obtain numbers below 1 - put any number into the function and you will not be able to numbers below 0.
to calculate the range, you simply find the inverse of the function and find the domain of the inverse function - or use differential calculus...
- #3
prasannapakkiam
But remember when finding the inverse of certain evil functions, that (SQRT(a))^2 or SQRT(a^2) = |a|, then set the other side of the equation to an inequality to RHS=>0. This trapped me a lot...
- #4
matt grime
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prasannapakkiam said:
or you could just write it down. Note in the case given, your method of finding the domain of the inverse won't work, since the domain is constrained by x>10, meaning the range is y>0. Your method yields y=/=0.
Just sketch the function, and look at it: it is always positive, and tends to infinty as x tends to 10 and 0 as x tends to infinity. It is continuous, hence the range is y>0.
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- #5
prasannapakkiam
well 'simply' "write" this one "down" in one step:
f(x)=(x^2-5x-9)^(1/4)
or
f(x)=x^x
f(x)=(x^2-5x-9)^(1/4)
or
f(x)=x^x
- #6
prasannapakkiam
which "case" are you referring to?
- #7
matt grime
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The case in the OPs link that he was asking about. You've not written down the domains of either of your functions, so it is impossible to write the ranges. Assuming you mean 'for the maximal subset of R for which this expression yields a real number' then the first is clearly y=>0. What you need to do depends on the question. The OPs question is straight foward to do by inspection.
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- #8
prasannapakkiam
matt grime: "Your method yields y=/=0"
I am sorry, but you probably need to finish solving this inequality (which I assume you did) - so finally my method yields:
y=/=0
y>0
y<=-1/10
Yes, I agree that my method does not couple the domain restriction of x>0. I have never encountered such a need to actually take into account a restiction before. This is one technique that I will find soon to implement upon this technique.
I am sorry, but you probably need to finish solving this inequality (which I assume you did) - so finally my method yields:
y=/=0
y>0
y<=-1/10
Yes, I agree that my method does not couple the domain restriction of x>0. I have never encountered such a need to actually take into account a restiction before. This is one technique that I will find soon to implement upon this technique.
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- #9
matt grime
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Oh, and your method also presupposes that the function is injective - it need not and in general will not be invertible.
- #10
prasannapakkiam
If by injection, you mean one-one - it does not matter. Read post #3. I mean, give one example through which my method completely breaks down...
Also (as I see it), the codomain is the set of all the Real values outputted by a function - thus this includes (unless stated otherwise), all the values of the function outputted by the domain.
Also (as I see it), the codomain is the set of all the Real values outputted by a function - thus this includes (unless stated otherwise), all the values of the function outputted by the domain.
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- #11
prasannapakkiam
Anyway Anived, remember post #3:
For example: f(x)=(x^2-5x-9)^(1/4)
To find the range, we can use my first method.
therefore: y=(x^2-5x-9)^(1/4)
therefore: y^4=|x^2-5x-9| -->
now LHS cannot be less than 0 as RHS is an absolute value.
thus we set:
y^4=>0
therefore: y=>0 ==>
now back to our first result:
y^4=|x^2-5x-9|
now there are 2 possible equations that can be made from this. However, in functions, each number in the domain can only have 1 answer. So we take the positive half:
therefore: y^4 = +(x^2-5x-9)
therefore: y^4 = (x-5/2)^2 -9-25/4
therefore: y^4 + 61/4 = (x-5/2)^2
therefore: SQRT(y^4 + 61/4)=|x-5/2|
now again set LHS => 0 (in this case it is unneccessary)
SQRT(y^4 + 61/4)>0
as you can see this yields yER==>
now back to the original equation - noting that again the positive bit is taken once again:
therefore: SQRT(y^4 + 61/4)=+(x-5/2)
therefore: SQRT(y^4 + 61/4)+5/2 = x
now the domain of this is also yER.==>
Now I finally write this collecting all the results:
Range = {y=>0}U{yER}U{yER}
thus
Range: {f(x)ER, f(x)=>0} ===>
Note: I think I should learn Latex...
For example: f(x)=(x^2-5x-9)^(1/4)
To find the range, we can use my first method.
therefore: y=(x^2-5x-9)^(1/4)
therefore: y^4=|x^2-5x-9| -->
now LHS cannot be less than 0 as RHS is an absolute value.
thus we set:
y^4=>0
therefore: y=>0 ==>
now back to our first result:
y^4=|x^2-5x-9|
now there are 2 possible equations that can be made from this. However, in functions, each number in the domain can only have 1 answer. So we take the positive half:
therefore: y^4 = +(x^2-5x-9)
therefore: y^4 = (x-5/2)^2 -9-25/4
therefore: y^4 + 61/4 = (x-5/2)^2
therefore: SQRT(y^4 + 61/4)=|x-5/2|
now again set LHS => 0 (in this case it is unneccessary)
SQRT(y^4 + 61/4)>0
as you can see this yields yER==>
now back to the original equation - noting that again the positive bit is taken once again:
therefore: SQRT(y^4 + 61/4)=+(x-5/2)
therefore: SQRT(y^4 + 61/4)+5/2 = x
now the domain of this is also yER.==>
Now I finally write this collecting all the results:
Range = {y=>0}U{yER}U{yER}
thus
Range: {f(x)ER, f(x)=>0} ===>
Note: I think I should learn Latex...
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