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(b) If f (c) = 0, the graph of f need not have an in ection point at x = c. For instance, the graph of f (x) = x 4 [see Fig. 23-4(b)] has a relative minimum, not an in ection point, at x = 0, where f (x) = 12x 2 = 0.

x!1 y!2 x!1 y!2

Fig. 23-4

Also, f 1; 2 12 2 2 5.

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23.2 TEST FOR RELATIVE EXTREMA We already know, from 14, that the condition f (c) = 0 is necessary, but not suf cient, for a differentiable function f to have a relative maximum or minimum at x = c. We need some additional information that will tell us whether a function actually has a relative extremum at a point where its derivative is zero. Theorem 23.3 (Second-Derivative Test for Relative Extrema): If f (c) = 0 and f (c) < 0, then f has a relative maximum at c. If f (c) = 0 and f (c) > 0, then f has a relative minimum at c. Proof: If f (c) = 0, the tangent line to the graph of f is horizontal at x = c. If, in addition, f (c) < 0, then, by Theorem 23.1,1 the graph of f is concave downward near x = c. Hence, near x = c, the graph of f must lie below the horizontal line through (c, f (c)); f thus has a relative maximum at x = c [see Fig. 23-5(a)]. A similar argument leads to a relative minimum when f (c) > 0 [see Fig. 23-5(b)].

By default, an AutoCAD VBA project contains a single module called ThisDrawing. You can then add UserForm, Standard and Class modules, and ActiveX controls as needed. 3 describes how to add these elements to an application. If you are an experienced VBA developer in AutoCAD 2000 or 2002, read the documentation in AutoCAD 2006 for new, changed, and removed features for ActiveX, VBA, system variables, and drawing data classes. Each new release brings considerable changes that often affect program development in good and bad ways.

Then lim f x; y f 1; 2 and the function is continuous at 1; 2 . Alternatively, we can show, in much the same manner as in the rst method of Problem 6.4, that given any > 0 we can nd > 0 such that j f x; y f 1; 2 j < when jx 1j < ; j y 2j < .

Fig. 23-5

2 x 2y; 6.6. Determine whether f x; y 0; (a) has a limit as x ! 1 and y ! 2,

order to use Theorem 23.1, we must assume that f is continuous at c and exists in an open interval around c. However, a more complicated argument can avoid that assumption.

x!1 y!2

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