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# Functions! Aaarrggghhh!!!!

Maths and statistics discussion, revision, exam and homework help.

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1. Functions! Aaarrggghhh!!!!
Okay, I have no idea where I'm going wrong with this!

If f(x) = sqrt(x+4) and the domain of f(x) = [0,infinity) then surely the range of f(x) would be [2,infinity)?

f(0) would be sqrt(0+4) = 2
f(infinity) would be sqrt(0+infinity) = infinity

The answer in the book, however is the range of f(x) = [1,infinity)

Why?

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2. Re: Functions! Aaarrggghhh!!!!
Your domain is wrong. Square roots are defined for positive numbers and zero; the argument then, can be 0 and greater. Therefore, for the smallest x in your domain you have x+4=0.

Just noticed that the book would still be wrong.. Were you given the domain to use? With the domain you have, you are correct.
Last edited by FireGarden; 24-07-2012 at 16:47.
3. Also, if g(x) = 2x^2 - 3
Domain: (-infinity, infinity)
Range: [-3, infinity)

If I were to find a value of x so that fg(x) cannot be formed, how would I go about finding it?

I'm really stuck, I checked the back of the book to see the answer, which is x= +/- sqrt(3/2) but I have no idea how they got that!

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4. (Original post by FireGarden)
Your domain is wrong. Square roots are defined for positive numbers and zero; the argument then, can be 0 and greater. Therefore, for the smallest x in your domain you have x+4=0.

Just noticed that the book would still be wrong.. Were you given the domain to use? With the domain you have, you are correct.
The domain is the one that was given to me in the question, how can it be wrong?

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5. Re: Functions! Aaarrggghhh!!!!
Well, technically it's not - part of defining a function is defining it's domain, but usually such questions ask you to 'find the domain', as in the interval for all the values that the function will work for. The interval they've given you is not the entire set of values that f can work with, but it's a valid function all the same.
6. As for the least value question, I would have thought that the composite function fg(x) would involve putting g(x) into f(x), including its domain, (-infinity,infinity) so surely a value of x that would mean the function couldn't form would be anything below 0 as f(x) domain is [0,infinity)

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7. Re: Functions! Aaarrggghhh!!!!
For fg(x), you'd be putting the range of g into f; so the domain of f in this composition would be [-3, infinity), but it's not , as f has its domain already defined; the composition does not affect this. So when you have g(x) = 0 (solve this, and you get the answer the book is telling you), you will then have to compute f(0), but by f's domain, this is 'unformed', as it's outside the range of allowed values. There are other values, too. Any x such that 0<g(x)<2 will also be unformed.

In short, the composition simply sticks the functions together, but part of defining a function is defining its domain - so this does not get messed with.
Last edited by FireGarden; 24-07-2012 at 17:11.
8. Thank you! But why do I need to put g(x) = 0?

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9. If any value of 0<g(x)<2 (not entirely sure where you got that from) leaves fg(x) unformed, then after solving for x, do I disregard x=-3/2 as it does not lie between 0 and 2?

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Last updated: July 24, 2012
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