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

APPLICATIONS OF THE SECOND DERIVATIVE AND GRAPH SKETCHING
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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