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0 = f (k) = an k n + an 1 k n 1 + + a1 k + a0 , then a0 = (an k n + an 1 k n 1 + + a1 k) = k(an k n 1 + an 1 k n 2 + + a1 ).

f (x)

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An array of doubles specifying a list of 3-D fit points in WCS coordinates. It s a simple array with a single dimension composed of alternating X, Y, and Z values (i.e., p1x, p1y, p1z, p2x, p2y, p2z, etc.). Because you must supply at least two points to create a spline, this array must have a minimum of six elements. A three-element array of doubles that determines the tangent of the spline at its first fit point. A three-element array of doubles that determines the tangent of the spline at its last fit point.

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as seems clear from the graph. To prove this we must show that given any > 0, we can nd > 0 such that j f x 1j < whenever 0 < x 1 < . Now since x > 1, f x 1 and so the proof consists in the triviality that j1 1j < whenever 0 < x 1 < . (c) As x ! 3 from the left, f x ! 1, i.e., lim f x 1. A proof can be formulated as in (b).

12.23 Let f (x) =

x!3

(a) Find the value of A for which f is continuous at x = 3. (b) For the value of A found in (a), is f differentiable at x = 3 1 12.24 Use the original de nition of derivative to nd the derivative of the function f such that f (x) = . x 1 12.25 Assuming that f is differentiable, prove: (a) If f is even, f is odd; (b) If f is odd, f is even.

Fig. 3-12

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DIFFERENTIABILITY AND CONTINUITY f (a + h) f (a) In the formula lim for the derivative f (a), we can let x = a + h and rewrite f (a) as h 0 h f (x) f (a) . If f is differentiable at a, then lim x a x a

The following code sample demonstrates the AddSpline method: Public Sub TestAddSpline() Dim objEnt As AcadSpline Dim dblBegin(0 To 2) As Double Dim dblEnd(0 To 2) As Double Dim dblPoints(14) As Double '' set tangencies dblBegin(0) = 1.5: dblBegin(1) = 0#: dblBegin(2) = 0 dblEnd(0) = 1.5: dblEnd(1) = 0#: dblEnd(2) = 0

x!3 x!3 x!3

lim f (x) = lim [f (x) f (a) + f (a)]

We must show that given any > 0, we can nd > 0 such that jx sin 1=x 0j < when 0 < jx 0j < . If 0 < jxj < , then jx sin 1=xj jxjj sin 1=xj @ jxj < since j sin 1=xj @ 1 for all x 6 0. Making the choice , we see that jx sin 1=xj < when 0 < jxj < , completing the proof.

= lim [f (x) f (a)] + lim f (a)

x!0 1

f (x) f (a) (x a) + f (a) = lim x a x a f (x) f (a) = lim lim (x a) + f (a) x a x a x a = f (a) 0 + f (a) = f (a) Thus, f is continuous at a. This proves: Theorem 13.1: If f is differentiable at a, then f is continuous at a. EXAMPLE Differentiability is a stronger condition than continuity. In other words, the converse of Theorem 13.1 is not true. To see this, consider the absolute-value function f (x) = |x| (see Fig. 13-1). f is obviously continuous at x = 0; but it is not differentiable at x = 0. In fact,

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