Find the Equation of the Tangent LineDate: 1/21/96 at 23:52:54 From: S. W. Davis Subject: Equation of the tangent line Find the the equation of the tangent line when x=1 f(x)= a/b a=2^x+1-sin(x+1/X+2) b=(x^3-2x+7)^1/3 Date: 7/13/96 at 22:43:43 From: Doctor Paul Subject: Re: Equation of the tangent line 2^x + 1 - sin((x+1)/(x+2) f(x)= ------------------------- (x^3-2x+7)^1/3 The first thing you must know is how to find the equation of a line: (y-y0) = m(x-x0) where m is the slope and (x0,y0) is a point on the line. We know what x0 is; that's given in the problem: it's 1. If we plug 1 in for x then we can get a number for f(x) and that will be out y0. That's not so hard. Then all we need is m. Let's first get y0. 2^1 + 1 - sin(2/3) f(x) = ------------------ (1^3-2(1)+7)^(1/3) 3 - .6183698031 = --------------- 1.817120593 = 1.310661607 so our formula looks like this: (y-1.310661607) = m*(x-1) To find m, we take the derivitive of f(x), and plug in x0 and y0 This derivitive is extremely messy..I used the quotient rule and here's what I got (broken up into two pieces..) : x x + 1 / 1 x + 1 \ 2 ln(2) - cos(-----) |----- - --------| x + 2 |x + 2 2| \ (x + 2) / -------------------------------- - 3 1/3 (x - 2 x + 7) / x x + 1 \ 2 |2 + 1 - sin(-----)| (3 x - 2) \ x + 2 / -------------------------------- 3 4/3 3*(x - 2x + 7) now plug in x = 1 and get the slope to be .6420381621 so the equation of the line is: (y-1.310661607) = .6420381621*(x-1) -Doctor Paul, The Math Forum |
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