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File0028 - Section 3.2 Differentiation Rules for...

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Unformatted text preview: Section 3.2 Differentiation Rules for Polynomials, Exponentials, Products, and Quotients 50. (a) y = x3 — 3x — 2 :> y’ 2 3x2 — 3. For the tangent to be horizontal, we need In : y’ z 0 :> 0 : 3x2 — 3 => 3X2 : 3 => x = :l: 1. When x : ~l, y : 0 :> the tangent line has equation y = O. The line perpendicular to this line at (—1,0) is x : —1. When x = 1, y : —4 :> the tangent line has equation y = —4. The line perpendicular to this line at (17 —4) is x = l. (b) The smallest value of y’ is -3, and this occurs when x = 0 and y = -—2. The tangent to the curve at (0, —2) has slope —3 => the line perpendicular to the tangent at (0, —2) has slope % => y + 2 = % (x — 0) or y : % x — 2 is an equation of the perpendicular line. _ 4x 91 _ (x2+l)(4)—(4x)(ZX) 4x2+4—8x2 4(—x2+1 _ _ r g «0+ 1) 51. y m —s dx (KM—5,— 0&2“)? _ (XZ+IJT1.thnx—0,y—Oandy _ 1 = 4, so the tangent to the curve at (0,0) is the line y : 4x. When x : 1, y z 2 => y’ = 0, so the tangent to the curve at (l, 2) is the line y = 2. _~ g _ x2+4)(0)—8(2x) _ —16 _ _ _ 716(2) _ 1 52, y m —\ y’ ( (X2+4)7— (“+13% When x —. 2, y _ l and y’ _ (2H4)? _ — 5, so the tangent line to the curve at (2, 1) has the equation y — l = — % (x — 2), or y : — g + 2. 53. y = ax2 + bx + 0 passes through (0,0) :> 0 = a(0) + b(0) T c :> c : O; y 2 3x2 + bx passes through (1, 2) :> 2 : a + b; y’ : 2ax + b and since the curve is tangent to y = x at the origin, its slope is 1 at x = 0 111 : y’:1whenx:0 =>1=2a(0)+b => b=1. Thena+b:2 :> a=l. Insummarya=b=landc=Oso the curve is y : x2 + x. 54. y=cx—x2 passes through (1,0) :> 0:c(1)—1 2 oz] :> the curve isyzx—xi’. For this curve, y’=1—-2xandx:1 :> y’: —l. Sincey=x—xzandyzx2+ax+bhavecommontangentsatx:O, y=x2+ax+bmustalsohaveslope~latx:1. Thusy’=2x+a => —1 :2-1+a => 21: ~3 => y = x2 —3x+b. Since thislast curve passes through (1,0), we haveO= l — 3+b :> b: 2. In summary, a:—3,b:2andc:—lsothecurvesarey:x2—3x+2andy:x—x2. 55. (a) y = x3 ~ x :5 y’ : 3x2 — 1. When x = —1, y : Oand y’ = 2 => the tangent line to the curve at(—1,0) is y=2(x+1)ory:2x+2i (b) _3 y—X—X ) y=2x+2 (c }=> x3—x=2x+2 => X3—3X—2=(X—2)(X+l)2=0 :> x:20rx:—1. Since y : 2(2) + 2 : 6; the other intersection point is (2, 6) 56. (a) y : x3 — 6x2 + 5x => y’ = 3x2 - 12x + 5. When x = 0, y = 0 and y’ = 5 :> the tangent line to the curve at (0,0) is y : 5x. ...
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