121616949-math.25

# 121616949-math.25 - 1.3 Functions 11 Here the box we get...

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1.3 Functions 11 Here the box we get will have height x and rectangular base of dimensions a - 2 x by b - 2 x . Thus, V = f ( x ) = x ( a - 2 x )( b - 2 x ) . Here a and b are constants, and V is the variable that depends on x , i.e., V is playing the role of y . This formula makes mathematical sense for any x , but in the story problem the domain is much less. In the ﬁrst place, x must be positive. In the second place, it must be less than half the length of either of the sides of the cardboard. Thus, the domain is 0 < x < 1 2 (minimum of a and b ) . In interval notation we write: the domain is the interval (0 , min( a, b ) / 2). EXAMPLE 1.4 Circle of radius r centered at the origin The equation for this circle is usually given in the form x 2 + y 2 = r 2 . To write the equation in the form y = f ( x ) we solve for y , obtaining y = ± r 2 - x 2 . But this is not a function , because when we substitute a value in ( - r, r ) for x there are two corresponding values of y . To get a function, we must choose one of the two signs in front of the square root. If we choose the positive sign, for example, we get the upper semicircle
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Unformatted text preview: y = f ( x ) = √ r 2-x 2 (see ﬁgure 1.6 ). The domain of this function is the interval [-r, r ], i.e., x must be between-r and r (including the endpoints). If x is outside of that interval, then r 2-x 2 is negative, and we cannot take the square root. In terms of the graph, this just means that there are no points on the curve whose x-coordinate is greater than r or less than-r .-r r . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1.6 Upper semicircle y = √ r 2-x 2...
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