Explanation Given information:- The given derivative of y = x 2 − x + x − 1 . Calculations:- The given derivative of y = x 2 − x + x − 1 . To find the transition point, interval of increase and decrease, concavity, and asymptotic behavior, we will take the first and second derivative of the given equation as shown below: y = x 2 − x + x − 1 [The given derivative of the equation] y ' = d d x [ x 2 − x + x − 1 . ] [Differentiate with respect to x ] y ' = d d x [ x 2 ] − d d x [ x ]+ d d x [ 1 x ] [Apply power rule [ u ( x ) n ] ' = n ⋅ u ( x ) n − 1 ] y ' = 2 x − 1 − d d x [ x ] x 2 [ ∴ , d d x [ x 2 ] = 2 x , d d x [ x ] = 1 ] y ' = 2 x − 1 − 1 x 2 Find the second derivative by using first derivative are as shown below: y ' = 2 x − 1 − 1 x 2 [Find the second derivative] y ' ' = d d x [ 2 x − 1 − 1 x 2 ] [Differentiate by the sum rule [ a ⋅ u ( x ) + b ⋅ v ( x ) ] ' = a ⋅ u ' ( x ) + b ⋅ v ' ( x ) ] y ' ' = 2 ⋅ d d x [ x ] − d d x [ 1 x 2 ] + d d x [ − 1 ] y ' ' = 2 x 3 + 2 For finding the transition point, the first derivative equal to zero and simplify the critical points

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ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 4.5, Problem 36E

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Chapter 4.5, Problem 35E

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Write the given third order linear equation as an equivalent system of first order equations with initial values. Use Y1 = Y, Y2 = y', and y3 = y". - - √ (3t¹ + 3 − t³)y" — y" + (3t² + 3)y' + (3t — 3t¹) y = 1 − 3t² \y(3) = 1, y′(3) = −2, y″(3) = −3 (8) - (888) - with initial values Y = If you don't get this in 3 tries, you can get a hint.

Question 2 1 pts Let A be the value of the triple integral SSS. (x³ y² z) dV where D is the region D bounded by the planes 3z + 5y = 15, 4z — 5y = 20, x = 0, x = 1, and z = 0. Then the value of sin(3A) is -0.003 0.496 -0.408 -0.420 0.384 -0.162 0.367 0.364

Chapter 4 Solutions

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