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Figure 11-7 Cartesian graph of the relation x2 y2 = 1
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The vertical-line test reveals that it isn t a function of x
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Consider the relation between an independent variable q and a dependent variable r defined by the equation
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r = 2q /p
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Sketch a graph of this equation in the polar plane Then redraw it in the Cartesian plane with q on the horizontal axis and r on the vertical axis Use the vertical-line test to determine, on the basis of the graph, whether or not the relation is a function of q
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Figure 11-8 is a graph of the equation in the polar plane It s a pair of dueling spirals When we draw the graph of the equation in a Cartesian plane with q on the horizontal axis and r on the vertical axis, we get a straight line that passes through the origin with a slope of 2 /p, as shown in Fig 11-9 The Cartesian vertical-line test indicates that the relation is a function of q
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Figure 11-8 Polar graph of the relation r = 2q /p Each
radial division represents 1 unit
Relations in Two-Space
r 3 2 1
2 3
Movable vertical line
Figure 11-9 Cartesian graph of the relation r = 2q /p
The vertical-line test shows that it s a function of q
Algebra with Functions
Functions can always be written as equations Therefore, when we want to add, subtract, multiply, or divide two functions, we can use ordinary algebra to add, subtract, multiply, or divide both sides of the equations representing the functions
Cautions There are three catches in the algebra of functions Whenever we add, subtract, multiply, or divide one function by another, we must watch out for these potential pitfalls Otherwise, we might get misleading or incorrect results:
The independent variables of the two functions must match That is, they must describe the same parameters or phenomena We can t algebraically combine functions of two different variables in an attempt to get a new function in a single variable If we try to do that, we won t know which variable the resultant function should operate on The domain of the resultant function is the intersection of the domains of the two functions we combine Any element in the domain of a sum, difference, product, or ratio function must belong to the domains of both of the constituent functions The domain of a ratio function may, however, be restricted even further if the denominator function becomes 0 anywhere in its domain
Algebra with Functions
If we divide a function by another function, the resultant function is undefined for any value of the independent variable where the denominator function becomes 0 This can, and often does, restrict the domain of the resultant function to a proper subset of the domain we would get if we were to add, subtract, or multiply the same two functions
New names for old functions So far in this chapter, we ve encountered four different functions Three of them are functions of x; the fourth is a function of q Following are the equations of the functions once again, for reference:
y=x 1 y = x2 y = x1/2 r = 2q /p Let s assign these functions specific names, so that we can write them in the conventional function notation These are f1 (x) = x 1 f2 (x) = x2 f3 (x) = x1/2 f4 (q) = 2q /p
Sum of two functions When we want to find the sum of two functions, we add both sides of their equations This can be done in either order, producing identical results For f1 and f2, we have
( f1 + f2)(x) = f1 (x) + f2 (x) = (x 1) + x2 = x2 + x 1 and ( f2 + f1)(x) = f2 (x) + f1 (x) = x2 + (x 1) = x2 + x 1 For f1 and f3, we have ( f1 + f3)(x) = f1 (x) + f3 (x) = (x 1) + x1/2 = x + x1/2 1 and ( f3 + f1)(x) = f3 (x) + f1 (x) = x1/2 + (x 1) = x + x1/2 1 For f2 and f3, we have ( f2 + f3)(x) = f2 (x) + f3 (x) = x2 + x1/2
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