Explanation Given information: The provided function is g ( x ) = ( x + 2 ) ( x 2 + x ) x − 4 and the point is ( 1 , − 2 ) . Formula used: According to the quotient rule of differentiation , the derivative of the quotient of the two differentiable functions is equal to the denominator with derivative of the numerator minus the numerator with derivative of the denominator and all divided by the square of the denominator. d d x [ f ( x ) g ( x ) ] = g ( x ) f ′ ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 , g ( x ) ≠ 0 The point slope form of the equation of the line is ( y − y 1 ) = m ( x − x 1 ) . Where, m is the slope of the line. Calculation: Consider the provided function is, g ( x ) = ( x + 2 ) ( x 2 + x ) x − 4 Simplify the above equation. g ( x ) = x 3 + x 2 + 2 x 2 + 2 x x − 4 g ( x ) = x 3 + 3 x 2 + 2 x x − 4 Apply the quotient rule for the derivative the function, g ( x ) = x 3 + 3 x 2 + 2 x x − 4 . g ′ ( x ) = ( x − 4 ) d d x ( x 3 + 3 x 2 + 2 x ) − ( x 3 + 3 x 2 + 2 x ) d d x ( x − 4 ) ( x − 4 ) 2 = ( x − 4 ) ( 3 x 2 + 6 x + 2 ) − ( x 3 + 3 x 2 + 2 x ) ( 1 ) ( x − 4 ) 2 = 3 x 3 − 12 x 2 + 6 x 2 − 24 x + 2 x − 8 − x 3 − 3 x 2 − 2 x ( x − 4 ) 2 = 2 x 3 − 9 x 2 − 24 x − 8 ( x − 4 ) 2 Now, substitute 1 for x in the above derivative to get the value of slope

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Author: Larson

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Chapter 2.4, Problem 54E

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To calculate: The equation of the tangent line to the graph of function g(x)=(x+2)(x2+x)x−4 at the point (1,−2).

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Chapter 2.4, Problem 53E

Chapter 2.4, Problem 55E

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Chapter 2 Solutions

CALCULUS: APPLIED APPROACH LOW COST MA

Ch. 2.1 - Checkpoint 1 Worked-out solution available at...Ch. 2.1 - Checkpoint2 Worked-out solution available at...Ch. 2.1 - Checkpoint 3 Worked-out solution available at...Ch. 2.1 - Checkpoint 4 Worked-out solution available at...Ch. 2.1 - Checkpoint 5 Worked-out solution available at...Ch. 2.1 - Checkpoint 6 Worked-out solution available at...Ch. 2.1 - Checkpoint 7 Worked-out solution available at...Ch. 2.1 - In Exercises 1-4, find an equation of the line...Ch. 2.1 - In Exercises 1-4, find an equation of the line...Ch. 2.1 - In Exercises 1-4, find an equation of the line...

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The equation of the tangent line to the graph of function g ( x ) = ( x + 2 ) ( x 2 + x ) x − 4 at the point ( 1 , − 2 ) .

Solution Summary: The author calculates the equation of the tangent line to the graph of function g(x) = left (x+2 )x-4 at the point

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To calculate: The equation of the tangent line to the graph of function g(x)=(x+2)(x2+x)x−4 at the point (1,−2).

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Chapter 2 Solutions