APPLICATIONS OF THE SECOND DERIVATIVE AND GRAPH SKETCHING in VS .NET

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APPLICATIONS OF THE SECOND DERIVATIVE AND GRAPH SKETCHING
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[CHAP. 23
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Since 1 x 2 = (1 x)(1 + x) , f (x) has a single positive root, x = 1, at which f (1) = [ 2(2)]/(2)3 = 1 . Hence, by the 2 second-derivative test, f has a relative maximum at x = 1. The maximum value is f (1) = 1 . 2 If we examine the formula for f (x) , 2x(x 3)(x + 3) f (x) = (x 2 + 1)3 we see that f (x) > 0 when x > 3 and that f (x) < 0 when 0 < x < 3. By Theorem 23.1, the graph of f is concave upward for x > 3 and concave downward for 0 < x < 3. Thus, there is an in ection point I at x = 3, where the concavity changes. Now calculate
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which shows that the positive x-axis is a horizontal asymptote to the right. The graph, with its extension to negative x (dashed), is sketched in Fig. 23-9. Note how concavity of one kind re ects into concavity of the other kind. Thus, there is an in ection point at x = 3 and another in ection point at x = 0. The value f (1) = 1 2 is the absolute maximum of f , and f ( 1) = 1 is the absolute minimum. 2
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Fig. 23-9
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Solved Problems
23.1 Sketch the graph of f (x) = x 1/x.
The function is odd. Hence, we can rst sketch the graph for x > 0 and, later, re ect in the origin to obtain the graph for x < 0. The rst and second derivatives are 1 f (x) = Dx (x x 1 ) = 1 ( 1)x 2 = 1 + 2 x 2 f (x) = Dx (1 + x 2 ) = 2x 3 = 3 x Since f (x) = 1 + (1/x 2 ) > 0, f is an increasing function. Moreover, for x > 0, the graph of f is concave downward, since f (x) = (2/x 3 ) < 0 when x > 0. The line y = x turns out to be an asymptote, because lim [x f (x)] = lim 1 =0
x +
x + x
CHAP. 23]
APPLICATIONS OF THE SECOND DERIVATIVE AND GRAPH SKETCHING
Since
x 0+
lim f (x) = lim
x 0+
= lim
x 0+
=
the graph of f has the negative y-axis as a vertical asymptote. Notice that x = 0, at which f is unde ned, is the only critical number. The graph is sketched, for all x, in Fig. 23-10. Although the concavity changes at x = 0, there is no in ection point there because f (0) is not de ned.
Fig. 23-10
Fig. 23-11
23.2 Sketch the graph of f (x) = 1 x 4 x 3 + 4x + 2. 4 The rst derivative is f (x) = x 3 3x 2 + 4. We can determine that 1 is a root of x 3 3x 2 + 4. algebra When looking for roots of a polynomial, rst test the integral factors of the constant. In this case, the factors of 4 are 1, 2, 4. So (Theorem 7.2), f (x) is divisible by x + 1. The division yields x 3 3x 2 + 4 = (x + 1)(x 2 4x + 4) = (x + 1)(x 2)2 Hence, the critical numbers are x = 1 and x = 2. Now f (x) = 3x 2 6x = 3x(x 2) Thus, f ( 1) = 3( 1)( 1 2) = 9. Hence, by the second-derivative test, f has a relative minimum at x = 1. Since f (2) = 3(2)(2 2) = 0, we do the rst-derivative test at x = 2. f (x) = (x + 1)(x 2)2 On both sides of x = 2, f (x) > 0, since x + 1 > 0 and (x 2)2 > 0. This is the case {+, +}. There is an in ection point at (2, 6). Furthermore, f (x) changes sign at x = 0, so that there is also an in ection point at (0, 2). Because
x
lim f (x) =
x 4
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