My Math Forum Range of a function

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 November 15th, 2009, 06:07 AM #1 Member   Joined: Oct 2009 Posts: 59 Thanks: 0 Range of a function Functions g and h are defined as follows: g : x ? 1 + x x ? R h : x ? x² + 2x x ? R Find i.) the ranges of g and h, range of g => R = {y : y ? R} 1 + x = 0 x = -1 1 - 2 = -1 -b/2a = -2/2 = -1 range of h => R = {y : y ? - 1, y ? R} ii.) the composite functions h o g and g o h, stating their ranges. Not sure how this is to be done help needed.
 November 15th, 2009, 07:31 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 The range of a function is the set of all values the function can have. It depends on the function's domain. Do you understand the term "domain"? For real x, 1 + x can have any real value, so its range is R, the set of all reals. For real x, since x² + 2x ?(x + 1)² - 1, its range is all real values ? -1. The composite function g(h(x)) ?1 + x² + 2x ?(x + 1)², so its range is all real values ? 0. What progress can you make in finding the range of the composite function h(g(x))?
November 15th, 2009, 08:47 AM   #3
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Re:

Quote:
 Originally Posted by skipjack The range of a function is the set of all values the function can have. It depends on the function's domain. Do you understand the term "domain"? For real x, 1 + x can have any real value, so its range is R, the set of all reals. For real x, since x² + 2x ?(x + 1)² - 1, its range is all real values ? -1.
Domain: The set of inputs for a function.
Represented mathematically it would be...R = {x : x ? R}
R = {x : x ? - 1, y ? R}

What progress can you make in finding the range of the composite function h(g(x))?
h{g(x)} ? h(1 + x) ? x² + 1 + 2(1 + x)²
? 3x² + 4x + 3.......I think?!

November 15th, 2009, 09:59 AM   #4
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Quote:
 Originally Posted by Tartarus Domain: The set of inputs for a function. Represented mathematically it would be...R = {x : x ? R}
Yes, although just R suffices; it's confusing to use R to refer to a domain, but perhaps you didn't intend that interpretation. The reals can alternatively be given as (-?, ?).

Quote:
 Originally Posted by Tartarus R = {x : x ? - 1, y ? R}
Use just one letter for the dummy variable, not two. As x ? - 1 implies x ? R, it suffices to give R = {x : x ? - 1}, or alternatively, R = [-1, ?).

Quote:
 Originally Posted by Tartarus h{g(x)} ? h(1 + x) ? x² + 1 + 2(1 + x)²
Not quite; h(1 + x) = (1 + x)² + 2(1 + x) = (1 + x)(3 + x) = x² + 4x + 3.
Can you obtain the range of this function?

November 15th, 2009, 11:14 AM   #5
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Re:

Quote:
Originally Posted by skipjack
Quote:
 Originally Posted by Tartarus Not quite; h(1 + x) = (1 + x)² + 2(1 + x) = (1 + x)(3 + x) = x² + 4x + 3. Can you obtain the range of this function?
h is (??, ?)

 November 15th, 2009, 09:49 PM #6 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Range of a function I think h(x) has a minimum at x = -2.
November 16th, 2009, 12:40 AM   #7
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Re: Range of a function

Quote:
 Originally Posted by aswoods I think h(x) has a minimum at x = -2.
So h is (?2, ?)?

 November 16th, 2009, 01:41 AM #8 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Range of a function The range will be [h(-2), ?) because the range is the output of h, not the input. So it's [-1, ?). By the way, are you supposed to use the convention of square brackets versus round brackets? So [-1... includes -1, but (-1... excludes it?
November 24th, 2009, 01:50 PM   #9
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Re: Range of a function

Quote:
 Originally Posted by aswoods The range will be [h(-2), ?) because the range is the output of h, not the input. So it's [-1, ?). By the way, are you supposed to use the convention of square brackets versus round brackets? So [-1... includes -1, but (-1... excludes it?
Thanks.
I'm not sure. The teacher never said anything about round or square brackets, although he uses the round ones.

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