Explanation Given: The function, f ( x ) = 1 − 1 x + 2 2 Concept Used: Slope of the tangent line of a function f ( x ) at any point x = c is given by the first derivative at that point. That is the slope of tangent line at x = c is given by f ′ ( c ) . The point-slope form of the line passing through the point ( x 1 , y 1 ) and having slope m is given by y − y 1 = m ( x − x 1 ) . Power rule of derivatives, d d x ( x n ) = n x n − 1 Quotient rule of derivatives, ( u v ) ′ = v ⋅ u ' − u ⋅ v ' v 2 Derivative of any constant C is zero, that is, d d x ( C ) = 0 Calculation: Consider the function, f ( x ) = 1 − 1 x + 2 2 In order to find the equation of tangent line, first we find the slope of tangent line by finding the first derivative at the specified point. So, using the quotient rule and power rule of derivatives, the first derivative of the function can be found as, f ' ( x ) = d d x ( 1 − 1 x + 2 2 ) = d d x ( 1 ) − d d x ( 1 x ) + d d x ( 2 2 ) = ( 0 ) − d d x ( x − 1 ) + ( 0 ) = − ( − 1 x − 1 − 1 ) = − ( −

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

6th Edition

ISBN: 9781524908102

Author: SMITH KARL J, STRAUSS MONTY J, TODA MAGDALENA DANIELE

Publisher: Kendall Hunt Publishing

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Chapter 3.2, Problem 32PS

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The equation of the tangent line to the graph of f(x)=1−1x+22 at x=4 .

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Chapter 3.2, Problem 31PS

Chapter 3.2, Problem 33PS

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Chapter 3 Solutions

Calculus: Special Edition: Chapters 1-5 (w/ WebAssign)

Ch. 3.1 - Prob. 1PS Ch. 3.1 - Prob. 2PS Ch. 3.1 - Prob. 3PS Ch. 3.1 - Prob. 4PS Ch. 3.1 - Prob. 5PS Ch. 3.1 - Prob. 6PS Ch. 3.1 - Prob. 7PS Ch. 3.1 - Prob. 8PS Ch. 3.1 - Prob. 9PS Ch. 3.1 - Prob. 10PS

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The equation of the tangent line to the graph of f ( x ) = 1 − 1 x + 2 2 at x = 4 .

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The equation of the tangent line to the graph of f(x)=1−1x+22 at x=4 .

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Chapter 3 Solutions