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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 49e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 3.1 - Problem 49e

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# Calculating derivatives f(x+h)f(x) a. or the following

ISBN: 9780321570567 2

## Solution for problem 49E Chapter 3.1

Calculus: Early Transcendentals | 1st Edition

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Problem 49E

Calculating derivatives f(x+h)?f(x) ? a. ? or the following functions, find f’ using the definitionf (x) = lh?0 h . ? b. ?Determine an equation of the line tangent to the g ? ra?p? o? f f at (?a,? )) ?for the given value of a. f(x) = 3x+1; a = 8

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SOLUTION STEP 1 (a). Given f(x) = 3x+1. We need to find the derivative of f(x) using the definition of the derivative f(x+h)f(x) The definition of the derivative is(x) = lim h h0 f(x+h) = 3x+h)+1 3(x+h)+1 3x+1 Thus f (x) = lim h h0 To solve this we have to multiply the numerator and the denominator with the conjugate of the numerator. 3(x+h)+1 3x+1 3(x+h)+1 3x+1 3(x+h)+1+ 3x+1 Therefore f(x) = lim h = lim h × 3(x+h)+1+ 3x+1 h0 h0 3(x+h)+1(3x+1) 3h =lh0 h 3(x+h)+1+ 3x+1h0 h(3(x+h)+1+ 3x+1 = 3 = 3 3x+1+ 3x+1 2 3x+1 Therefore f (x) = 3 2 3x+1 STEP 2 (b). Determine an equation of the line tangent to the graph of f at (a, f(a)),{ie,(8,f(8))} for the given value a=8. Equation of a tangent line at (a,f(a)) is y y = m(x x ) 1 1 3 3 3 3 Slope is the m here which is given by the equation m = f (a) = f(8) = 2 3(8)+1= 2 25= 2×5 = 10 f(a) = f(8) =3(8)+1 = 24+1 = 25 = 5 Therefore the line tangent to the graph of f at (8,5) for the given value a=8 is 3 3 24 y5 = 10(x8) = 10x 10 y = 3x 24+5 = 3x 26 10 10 10 10 The equation of the line is y = 3x 26 10 10

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