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Domain and range question

  • Thread starter joejo
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  • #1
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domain and range question!!

Hi,

I have a quick domain and range question, I have placed my answer below the question. I'm having doubts that it is right! Can someone please help me verify it. Thanks in advance!
_____
Let f(x)= x² + 2 and g(x) =√1-x²

a) Find the domain and range of f and g.
b) Are the functions f.g and g.f defined? Explain?

ANSWER:

The domain of f is {X|X ε R}
The range of f is {y|y ≥ 2, yε R}

f.g= f[g(x)]
= f ___
(√ 1-x²)
= ___
(√ 1-x²) + 2
= 1- x² + 2
= 3-x²

All of the above is correct. I have also already done g.f as well.

This is where i need help....what comes next....
Are the functions f.g and g.f defined? Explain? and what is the domain of g and the range of g?

Thanks in advance for your help!!
 
Last edited:

Answers and Replies

  • #2
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They are defined, because they both exist.

The domain of g is all numbers less than and including 1, because once you get alrger than one, you will be taking the square root of negatives.
 
  • #3
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hey thanks for the quick reply, but how about the range of g?
 
  • #4
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I gave you a huge hint on the lower limit of the range of g, and knowing the domain of g you can figure out the lupper limit on the range of g.
 
  • #5
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well this is what i have so far,

the doman of g is {x||x| ≤ + 1}
the range of g is {y|y ≥ 0 ≤ y ≤ 1}

g of f isn't defined because domain is not real?!!

is that right
 
  • #6
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joejo said:
well this is what i have so far,

the doman of g is {x||x| ≤ + 1}
It doesnt need to be absolute x if your squaring it.

the range of g is {y|y ≥ 0 ≤ y ≤ 1}
Correct
g of f isn't defined because domain is not real?!!

is that right
[tex]g=\sqrt{1-x^2} [/tex]

[tex] f=x^2+2 [/tex]

[tex] g(f(x)) = \sqrt{1-(x^2+2)^2} [/tex]

So (x^2+2)^2 must be less than 1, work from here.
 
  • #7
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sorry you're losing me...ive been trying for the past 15 minutes....anyone else or whozum can you please clairfy
 
Last edited:
  • #8
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It doesnt need to be absolute x if your squaring it.
Actually it does... precisely because you are squaring it! Look again :tongue:

And while "range of [itex]g = \{ y | y \geq 0 \leq y \leq 1 \}[/itex]" is correct, it's also a pretty strange way of writing it. You can just write [itex]\{ y | 0 \leq y \leq 1\}[/itex].
 
  • #9
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What are you missing
 
  • #10
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You are quite right that [itex]g \circ f[/itex] isn't defined over [itex]\mathbb{R}[/itex] for any real [itex]x[/itex].

Whozum's anaylsis was right, but then again so was your comment he responded to: [itex](x^2 + 2)^2 \leq 1[/itex] is not satisfied for any real [itex]x[/itex].
 
  • #11
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so is this right?

the domain of g is {x||x| ≤ + 1}
 
  • #12
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indeed it is.
 
  • #13
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so it's not.......

the domain of g is {x| x ≤ + 1}
 
  • #14
998
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well, is [itex]g(-2)[/itex] defined? Is [itex]-2 \leq 1[/itex]? That should answer your question! :tongue:
 
  • #15
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x^2 is always positive. saying |x|^2 is redundant, isnt it.
 
  • #16
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no its not!
 
  • #17
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data no it isn't....
 
  • #18
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Oh ok. I see what you mean by absolute. I was thinking x can go from -1,1 which is the same as |x| from 0 to 1
 
  • #19
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Yes it is. But precisely because of that,

[tex]g(x) = \sqrt{1 - x^2}[/tex]

is real if and only if [itex] |x| < 1[/itex]. If [itex]|x| > 1[/itex] then either [itex] x < -1[/itex] or [itex] x > 1[/itex]. In either case, [itex](-x)^2 = x^2 > 1[/itex] so [itex] 1 < x^2 \Longrightarrow 1 - x^2 < 0[/itex].
 
  • #20
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how come your latex looks so sexy
 
  • #21
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I was thinking x can go from -1,1 which is the same as |x| from 0 to 1
precisely. And [itex] 0 \leq |x| \leq 1[/itex] is the same as [itex]|x| \leq 1[/itex] by the definition of an inner product.
 
  • #22
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how come your latex looks so sexy
Long physics lab reports. Believe me, it's not a good tradeoff :smile:
 
  • #23
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I think Im dernaged because I yearn those.
 
  • #24
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hey data....

can't i just say the domain of g is ....

{x|x ≤ + 1}
 
  • #25
998
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Once you spend two years of friday nights writing them for twelve hours in a row (like me!), you might not be so enamored with the idea anymore :rofl:
 

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