37_Cal_Solution of Calculus_6e

# 37_Cal_Solution of Calculus_6e - y(5 = 1(x 0 2...

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y ( 5) = 1 2 ( x 0); alternatively, 2 y + 10 = x. C01S0M.022: The equation 3 x 2 y = 4 of the other line may be written in the form y = 3 2 x 2, revealing that it and L have slope 3 2 . Hence an equation of L is y ( 3) = 3 2 ( x 2); that is, y = 3 2 x 6 . C01S0M.023: The equation y 2 x = 10 may be written in the form y = 2 x + 10, showing that it has slope 2. Hence the perpendicular line L has slope 1 2 . Therefore an equation of L is y 7 = 1 2 ( x ( 3)); that is, x + 2 y = 11 . C01S0M.024: The segment S joining (1 , 5) and (3 , 1) has slope ( 1 ( 5)) / (3 1) = 2 and midpoint (2 , 3), and hence L has slope 1 2 and passes through (2 , 3). So an equation of L is y ( 3) = 1 2 ( x 2); that is, x + 2 y = 4 . C01S0M.025: The graph of y = f ( x ) = 2 2 x x 2 is a parabola opening downward. The only such graph is shown in Fig. 1.MP.6. C01S0M.026: Given: f ( x ) = x 3 4 x 2 + 5. Because f ( 1) = 0, f (1) = 2 > 0 > 3 = f (2), and f (3) = 4 < 0 < 5 = f (4), the graph of f crosses the x -axis at x = 1, between x = 1 and x = 2, and between x = 3 and x = 4. Hence the graph of
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