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Chapter 4 91

# Chapter 4 91 - SECTION 4.6 GRAPHING WITH CALCULUS AND...

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Unformatted text preview: SECTION 4.6 GRAPHING WITH CALCULUS AND CALCULATORS 353 33. f(a:) 2 cm + since : f'(m) 2 0+ coszz: é f”(at) : — sinm f(—:c) = —f(z). so f is an odd function and its graph is symmetric with respect to the origin. f(a:) = 0 41> sins: = —c;v. so 0 is always an :c—intercept. f/(zv) : 0 4:) cosm = —c. so there is no critical number when |c| > 1. If |c| S 1. then there are inﬁnitely many critical numbers. [f w] is the unique solution of cos a: = —c in the interval [0. 7r]. then the critical numbers are 34. 2n7r j: 201. where n ranges over the integers. (Special cases: When c = 1. m1 = 0: when 0 = 0. m = g; and when 0 = —1. \$1 = 7r.) f”(m) < 0 (i) sinm > 0. so f is CD on intervals of the form (27m, (2n —|— 1)7r). f is CU on intervals of the form ((271 i 1)7r. 27m). The inﬂection points of f are the points (27m. 2mm). where n is an integer. lfc Z 1. then f’(a:) Z 0 for all m. so f is increasing and has no extremum. Ifc S —1. then f’(av) S 0 for all cc. so f is decreasing and has no extremum. [f |cl < 1. then f’(:c) > 0 4:) cosar: > —c c) as is in an interval of the form (27m — \$1. 21m + x1) for some integer 71. These are the intervals on which f is increasing. Similarly, we ﬁnd that f is decreasing on the intervals of the form (2mr + x1. 2(n + 1)7r — an). Thus. f has local maxima at the points 2n7r + 331. where f has the values C(2mr + m) —l— sin m1 = C(2mr + \$1) + m. and f has local minima at the points 2mr — 331. where we have f(2n7r — m1) : C(2n7r — x1) — sin m1 : C(2mr i 321) i m. The transitional values of c are 71 and 1. The inﬂection points move vertically, but not horizontally. when 6 changes. When |c[ 2 1. there is no extremum. For |c| < 1. the maxima are spaced 27r apart horizontally. as are the minima. The horizontal spacing between maxima and adjacent minima is regular (and equals 7r) when c 2 0. but the horizontal space between a local maximum and the nearest local minimum shrinks as |c| approaches 1. For f(t) = C(e’” — 6—“). C affects only vertical stretching. so we let C : 1. From the ﬁrst ﬁgure, we notice that the graphs all pass through the origin. approach the t—axis as It increases. and approach —00 as t —> —00. Next we let a : 2 and produce the second ﬁgure. Here. as 1) increases. the slope of the tangent at the origin increases and the local maximum value increases. f0?) : 8—21: * 6’1” => f'(t) : be‘bt ~ 26—2‘. f’(0) : b — 2. which increases as b increases. ...
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