is not a function. It is not an injection because it’s not one-to-one. That means it cannot be a
bijection. The relation does map onto its entire range (the set of all reals), so it’s a surjection.Chapter 14
- If we multiply x by −1 and leave y the same, the point will move to the other side of the
y axis, but it will stay on the same side of the x axis. If it starts out in the first quadrant,
it will move to the second. If it starts out in the second quadrant, it will move to the
first. If it starts out in the third quadrant, it will move to the fourth. If it starts out in
the fourth quadrant, it will move to the third. The y axis will act as a “point reflector.” - If we multiply y by −1 and leave x the same, the point will move to the other side of the
x axis, but it will stay on the same side of the y axis. If it starts out in the first quadrant,
it will move to the fourth. If it starts out in the second quadrant, it will move to the
third. If it starts out in the third quadrant, it will move to the second. If it starts out in
the fourth quadrant, it will move to the first. The x axis will act as a “point reflector.” - The point for (6x, 6y) will be in the same quadrant as the point for (x,y), but 6 times
as far from the origin. The point for (x/4,y/4) will be in the same quadrant as the point
for (x,y), but 1/4 of the distance from the origin. The origin, the point for (x,y), the
point for (6x, 6y), and the point for (x/4,y/4) will all lie along a single straight line. - If the vertical test line intersects the graph (once for a function, and once or more
for a relation), then the point where the test line intersects the independent-variable
(horizontal) axis represents a numerical value in the domain. If the test line does not
intersect the graph, then the point where the test line intersects the horizontal axis does
not represent a value in the domain. - This process works just like the process for determining whether or not a point is in the
domain, except that everything is rotated by 90°! If the horizontal test line intersects the
graph, then the point where the test line intersects the dependent-variable (vertical) axis
represents a numerical value in the range. If the test line does not intersect the graph,
then the point where the test line intersects the vertical axis does not represent a value
in the range. - We can plot several specific points for y= |x|, and then we can determine the graph on
the basis of those points. We can deduce, using some common sense, that the lines are
straight. See Fig. B-1. The vertical-line test tells us that this is a function of x. - We can plot several specific points for y= |x+ 1|, and then we can determine the graph
on that basis. Again, we can deduce, using some common sense, that the lines are
straight. See Fig. B-2. The vertical-line test reveals that this is a function of x. - Figure B-3 is a graph of the inverse of y=x+ 1. If we apply the “point reflector”
method to Fig. 14-11, we don’t have to mathematically derive an equation for the
inverse to figure out what its graph looks like. - Figure B-4 is a graph of the inverse of w=v^2. This is the result of using the “point
reflector” scheme to modify Fig. 14-12. Again, it is not necessary to derive an equation
for the inverse.
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