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November 15th, 2009, 06:07 AM   #1
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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.
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November 15th, 2009, 07:31 AM   #2
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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))?
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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 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?!
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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) ? 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?
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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(1 + x) = (1 + x)(3 + x) = x + 4x + 3.
Can you obtain the range of this function?
h is (??, ?)
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November 15th, 2009, 09:49 PM   #6
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Re: Range of a function

I think h(x) has a minimum at x = -2.
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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, ?)?
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November 16th, 2009, 01:41 AM   #8
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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?
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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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