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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 ? x˛ + 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: 20,919 Thanks: 2203 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 x˛ + 2x ?(x + 1)˛ - 1, its range is all real values ? -1. The composite function g(h(x)) ?1 + x˛ + 2x ?(x + 1)˛, 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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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 x˛ + 2x ?(x + 1)˛ - 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) ? x˛ + 1 + 2(1 + x)˛
? 3x˛ + 4x + 3.......I think?! November 15th, 2009, 09:59 AM   #4
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 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) ? x˛ + 1 + 2(1 + x)˛
Not quite; h(1 + x) = (1 + x)˛ + 2(1 + x) = (1 + x)(3 + x) = x˛ + 4x + 3.
Can you obtain the range of this function? November 15th, 2009, 11:14 AM   #5
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Originally Posted by skipjack
Quote:
 Originally Posted by Tartarus Not quite; h(1 + x) = (1 + x)˛ + 2(1 + x) = (1 + x)(3 + x) = x˛ + 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. Tags function, range Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post th3j35t3r Number Theory 5 November 17th, 2013 01:47 PM happy21 Calculus 9 August 17th, 2013 07:31 AM sachinrajsharma Calculus 2 April 8th, 2013 10:09 PM balbasur Algebra 1 May 30th, 2010 04:28 PM hobilla Algebra 3 March 13th, 2008 08:20 PM

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