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By Stanley I Grossman

ISBN-10: 0123043719

ISBN-13: 9780123043719

ISBN-10: 012304376X

ISBN-13: 9780123043764

1,178 pages plus Appendixes of 146 pages

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Extra resources for Calculus

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Letf(x) = vX+l and g(x) = \/4'=X2. Since dom f = [ - 1, oo) and domg = [ - 2, 2], we have dom( f + g) = dom( f - g) = dom( f· g) = [ - 1, oo) n [ - 2, 2] = [ - 1, 2], and dom(flg) = [ - 1, 2] - {x : � = O} = [ - 1, 2] - { - 2, 2} = [ - 1, 2). The func­ tions are 1 . 7 01'1RA110:'-JS \\'I flI I U'\:L I IONS 45 ( f + g)(x) = vX+1 + � ( f - g)(x) = vX+1 - � ( f g ) (x) = vX+J . � = v'(x + 1)(4 - x 2) • (f) (x) g = vX+1 = v'4 - x 2 �x + 1 . 4 - x2 • Definition 1 can lead to some confusion if we are not careful.

Note, for example, that (f 0 g )(3) = v1o, while (g dom g = IR, and we have+ dom(f 0 g) = {x : g (x) = x2 + 1 0 f)(3) = 4 . Now dom f = IR + , E dom f}. But since x2 + 1 > 0, x 2 + 1 E dom f for every real x, so dom(f g) = IR. On the other hand, dom(g 0 f) = IR + since f is only defined for x ::::: 0. • 0 • EXAMPLE 4 WARNING As Example 3 suggests, it is Let f(x) = 3x - 4 and g(x) = x3• Then not true in general that (f 0 g)(x) = (g 0 f)(x). (f g)(x) = f(g(x)) = f(x3) = 3x3 - 4 0 and (g 0 f)(x) = g(f(x)) = g (3x - 4) = (3x - 4)3• Here dom(f 0 g) = dom(g 0 f) = IR.

This circle is sketched in Figure 7a. • In general, the equation of the circle with center at (a, b ) and radius r is given by (x - a )2 + (y - b )2 = r 2 • (4) This circle is sketched in Figure 7b. EXAMPLE 7 Show that the equation x 2 - 6x + y2 + 2y - 17 = center and radius. 0 represents a circle, and find its 18 C lI . I l'Rt1 IMINARIES v v (al (b) FIGURE 7 Solution. + We have x2 - 6x + y2 + 2y - 17 = (x 2 - 6x + 9) + (y2 + 2y + 1) = (x - 3)2 + (y + 1 )2 - 27 = 0. - 9 - 1 - 17 This result implies that (x - 3)2 + (y + 1)2 = 27, which is the equation of a circle with center at (3, - 1) and radius VV.

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Calculus by Stanley I Grossman


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