Csi Algebra 2 and Pre Calc Families of Functions Answers

Functions: Domain and Range

Let's return to the field of study of domains and ranges.

When functions are first introduced, y'all will probably take some simplistic "functions" and relations to deal with, usually being merely sets of points. These won't be terribly useful or interesting functions and relations, only your text wants you to go the thought of what the domain and range of a function are. Small sets of points are generally the simplest sorts of relations, and then your volume starts with those.

For example:

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• State the domain and range of the following relation. Is the relation a function?

{(two, –three), (iv, half-dozen), (3, –ane), (half-dozen, half-dozen), (2, three)}

The to a higher place list of points, being a relationship betwixt certain ten 's and sure y 's, is a relation. The domain is all the x -values, and the range is all the y -values. To give the domain and the range, I just list the values without duplication:

domain: {2, 3, 4, six}

range: {–3, –one, 3, 6}

(It is customary to list these values in numerical lodge, but it is not required. Sets are called "unordered lists", so you can listing the numbers in whatever order you feel similar. Just don't indistinguishable: technically, repetitions are okay in sets, only about instructors would count off for this.)

While the given set does indeed represent a relation (because x 's and y 's are beingness related to each other), the ready they gave me contains 2 points with the same ten -value: (2, –iii) and (two, 3). Since ten = 2 gives me 2 possible destinations (that is, two possible y -values), and then this relation is not a office.

Note that all I had to do to check whether the relation was a function was to look for duplicate x -values. If you find any duplicate 10 -values, then the different y -values mean that you do non have a function. Recall: For a relation to be a function, each ten-value has to go to one, and only 1, y-value.

• State the domain and range of the following relation. Is the relation a function?

{(–3, v), (–2, v), (–ane, 5), (0, 5), (one, five), (2, 5)}

I'll just listing the ten -values for the domain and the y -values for the range:

domain: {–iii, –two, –1, 0, 1, 2}

range: {5}

This is some other instance of a "boring" function, simply similar the example on the previous page: every concluding x -value goes to the exact same y -value. But each x -value is dissimilar, and so, while irksome,

this relation is indeed a function.

In point of fact, these points lie on the horizontal line y = 5.

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Past the fashion, the name for a fix with only one element in information technology, like the "range" set above, is "singleton". So the range could likewise exist stated as "the singleton of five"

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There is ane other instance for finding the domain and range of functions. They will give y'all a function and ask you to find the domain (and perchance the range, likewise). I have only e'er seen (or can fifty-fifty think of) two things at this stage in your mathematical career that y'all'll have to check in social club to decide the domain of the role they'll give you, and those ii things are denominators and square roots.

• Make up one's mind the domain and range of the given function:

The domain is all the values that ten is allowed to take on. The only problem I take with this function is that I need to be careful not to divide by nil. So the only values that x can not take on are those which would cause division by naught. And then I'll set the denominator equal to zero and solve; my domain volition exist everything else.

10 ² – ten – 2 = 0

(x – ii)(x + 1) = 0

ten = 2 or x = –i

Then the domain is "all x not equal to –ane or two".

The range is a scrap trickier, which is why they may not ask for it. In general, though, they'll want you to graph the office and find the range from the moving-picture show. In this case:

Equally you can run across from my picture, the graph "covers" all y -values; that is, the graph volition go as low every bit I similar, and will too go as high equally I like. For any betoken on the y -centrality, no matter how loftier up or low down, I tin go from that betoken either to the correct or to the left and, eventually, I'll cross the graph. Since the graph will eventually encompass all possible values of y , then:

the range is "all real numbers".

• Determine the domain and range of the given part:

The domain is all values that x can take on. The only trouble I have with this function is that I cannot have a negative within the square root. So I'll set the insides greater-than-or-equal-to zero, and solve. The upshot will be my domain:

–2ten + iii ≥ 0

–twox ≥ –3

2x ≤ 3

x ≤ 3/two = 1.5

Then the domain is "all x ≤ iii/ii".

The range requires a graph. I need to be careful when graphing radicals:

The graph starts at y = 0 and goes down (heading to the left) from there. While the graph goes downwardly very slowly, I know that, somewhen, I can go as depression every bit I similar (past picking an x that is sufficiently big). Also, from my feel with graphing, I know that the graph will never offset coming back up. Then:

the range is "all y ≤ 0".

• Determine the domain and range of the given function:

y = –x ⁴ + four

This is but a garden-variety polynomial. There are no denominators (and so no segmentation-by-aught issues) and no radicals (so no foursquare-root-of-a-negative problems). At that place are no problems with a polynomial. There are no values that I can't plug in for x . When I have a polynomial, the answer for the domain is always:

the domain is "all 10 ".

The range will vary from polynomial to polynomial, and they probably won't even ask, but when they exercise, I look at the picture:

The graph goes only as high as y = iv, but it will go as depression as I similar. And then:

The range is "all y ≤ four".

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Source: https://www.purplemath.com/modules/fcns2.htm

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