Explanation Given information:- The given derivative of f ( x ) = Sin 2 x − 4 cos x . Calculations:- The given derivative of f ( x ) = Sin 2 x − 4 cos x . Find the inflection point by using the first and second derivative of the given equation as shown below: f ( x ) = Sin 2 x − 4 cos x [The given derivative of the equation] y ' = d d x [ Sin 2 x − 4 cos x ] [Differentiate with respect to x ] y ' = [ d d x [ Sin 2 x ] − 4 ⋅ d d x [ cos ( x ) ] ] [where, sin x = cos x and cos x = − sin x ] y ' = 2 ⋅ cos ( 2 x ) + 4 sin x Find the second derivative by using first derivative are as shown below: y ' = 2 ⋅ cos ( 2 x ) + 4 sin x [Find the second derivative] y ' ' = d d x [ 2 ⋅ cos

4th Edition

ISBN: 9781319434526

Author: Rogawski

Publisher: MAC HIGHER

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Limits

When it comes to calculus, limits are considered to be a very important topic of discussion. The main importance of limit lies in expressing the value of a function as it gets closer to a certain value. Also, limits can help you to understand other topic…

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Chapter 4, Problem 44CRE

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The solution of point of inflection.

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Chapter 4, Problem 43CRE

Chapter 4, Problem 45CRE

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3.1 Limits 1. If lim f(x)=-6 and lim f(x)=5, then lim f(x). Explain your choice. x+3° x+3* x+3 (a) Is 5 (c) Does not exist (b) is 6 (d) is infinite

1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .

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Chapter 4 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

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The solution of point of inflection.

Solution Summary: The author calculates the inflection point by using the first and second derivatives of the given equation.

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Chapter 4 Solutions