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## Main Question or Discussion Point

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. Can someone please explain? Thanks in advance.

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- #2

prasannapakkiam

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

- #4

matt grime

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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.to calculate the range, you simply find the inverse of the function and find the domain of the inverse function - or use differential calculus...

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

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matt grime

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- #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 completly 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

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 (

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} ===>

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