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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 51e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 51e

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Calculating derivatives a. or the following functions,

ISBN: 9780321570567 2

Solution for problem 51E Chapter 3.1

Calculus: Early Transcendentals | 1st Edition

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Problem 51E

Calculating derivatives ? a. ? or the following functions, find f using the de ? finition f (x) = lim f(x+hh?f(x). h?0 ? b. ?Determine an equation of the line tangent to the gr?ap?h?of? f at (?a, ?f(a ? )) ?for the given value of a. 2 f(x) = 3x+1 ;a=-1

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SOLUTION STEP 1 (a). Given f(x) = 2 . 3x+1 We need to find the derivative of f(x) using the definition of the derivative The definition of the derivative is f (x) = lim f(x+h)f(x) h0 h 2 f(x+h) = 3(x+h)+1 2 2 2(3x+3h+1)(3x+1) Thus f (x) = lim 3x+3hh13x+1= lim h = lim h(3x+3h+1)(3x+1) h0 h0 h0 = lim 6 = 6 2 h0 (3x+3h+1)(3x+1) (3x+1) STEP 2 (b). Determine an equation of the line tangent to the graph of f at (a, f(a)) ,{ie,(-1,f(-1))} for the given value a=-1. Equation of a tangent line at (a,f(a)) is y y = m(x x ) 1 1 6 6 6 3 Slope is the m here which is given by the equation m = f (a) = f(1) = (3(1)+1)= 22= 4 = 2 f(a) = f(1) = 3(1)+1=1 Therefore the line tangent to the graph of f at (-1,-1) for the given value a=-1 is 3 3 3 y(1) = 2(x(1)) = 2x 2 3 3 y+1 = 2 x 2 y == 3x 1 == 3x 5 2 2 2 2 The equation of the line is 3 5 y == 2 x 2

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