Explanation Given Information: The provided equation of the curve is, y e x = 2 y + 1 and the point is ( 0 , − 1 ) . Formula used: The slope of tangent to a curve y = f ( x ) at point ( x , y ) is given by the derivative of the curve at that point. The equation of a line passing through points ( x 1 , y 1 ) and slope m is given by, y − y 1 = m ( x − x 1 ) Calculation: Consider the provided equation of curve, y e x = 2 y + 1 Now, find the derivative d y d x from y e x = 2 y + 1 by take the derivative term by term on both sides of the equation as, d d x ( y e x ) = d d x ( 2 y ) + d d x ( 1 ) y e x + e x d y d x = 2 d y d x Solve for d y d x as, y e x + e x d y d x = 2 d y d x y e x = 2 d y d x − e x d y d x <

Mathematical Applications for the Management, Life, and Social Sciences

11th Edition

ISBN: 9781305108042

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

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Chapter 11.3, Problem 46E

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To calculate: The equation of the tangent line to the curve yex=2y+1 at (0,−1).

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Chapter 11.3, Problem 45E

Chapter 11.3, Problem 47E

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Chapter 11 Solutions

Mathematical Applications for the Management, Life, and Social Sciences

Ch. 11.1 - 1. Ch. 11.1 - 2. If Ch. 11.1 - Prob. 3CP Ch. 11.1 - 4. Find . Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E Ch. 11.1 - Prob. 4E Ch. 11.1 - Prob. 5E Ch. 11.1 - Find the derivatives of the functions in Problems...

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The equation of the tangent line to the curve y e x = 2 y + 1 at ( 0 , − 1 ) .

Solution Summary: The author calculates the equation of the tangent line to the curve, yex=2y+1, and the slope of a line passing through points.

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To calculate: The equation of the tangent line to the curve yex=2y+1 at (0,−1).

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Chapter 11 Solutions