Explanation Given information: The given function is, f ( x ) = x 3 + x + 2 2 x 2 − 5 x + 4 . The value of x is 1.2 . Definition used: The first derivative of the function is the differentiation of the function, f ′ ( x ) = d d x f ( x ) (1) The second derivative of the function is the differentiation of the first derivative of the function, f ″ ( x ) = d d x f ′ ( x ) (2) The chain rule stated that the derivation of the function, f ( g ( x ) ) is f ′ ( g ( x ) ) ⋅ g ′ ( x ) . Formula used: The power rule is, d d x a x n = a ⋅ n ⋅ x n − 1 The quotient rule of the derivative is, d d x ( u v ) = v d u d x − u d v d x v 2 Calculation: Substitute x 3 + x + 2 2 x 2 − 5 x + 4 for f ( x ) in equation (1) to find the first derivative of the function, f ′ ( x ) = d d x [ x 3 + x + 2 2 x 2 − 5 x + 4 ] = ( 2 x 2 − 5 x + 4 ) d d x ( x 3 + x + 2 ) − ( x 3 + x + 2 ) d d x ( 2 x 2 − 5 x + 4 ) ( 2 x 2 − 5 x + 4 ) 2 ( Use qoutient rule ) = ( 2 x 2 − 5 x + 4 ) ( 3 x 2 + 1 ) − ( x 3 + x + 2 ) ( 4 x − 5 ) ( 2 x 2 − 5 x + 4 ) 2 ( Use power rule ) = ( 6 x 4 − 15 x 3 + 12 x 2 + 2 x 2 − 5 x + 4 ) − ( 4 x 4 + 4 x 2 + 8 x − 5 x 3 − 5 x − 10 ) ( 2 x 2 − 5 x + 4 ) 2 Further the simplification of the above equation is, f ′ ( x ) = 6 x 4 − 15 x 3 + 12 x 2 + 2 x 2 − 5 x + 4 − 4 x 4 − 4 x 2 − 8 x + 5 x 3 + 5 x + 10 ( 2 x 2 − 5 x + 4 ) 2 = 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 Substitute 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 for f ′ ( x ) in equation (2) to find the second derivative of the function, f ″ ( x ) = d d x [ 2 x 4 − 10 x 3 + 10 x 2 − 8 x + 14 ( 2 x 2 − 5 x + 4 ) 2 ] = ( 2

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

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ISBN: 9780357465400

Author: Tan

Publisher: CENGAGE L

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Chapter 3.5, Problem 6TE

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The value of second derivative of the function at given value of x.

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Each of the following statements is an attempt to show that a given series is convergent or divergent using the Comparison Test (NOT the Limit Comparison Test.) For each statement, enter C (for "correct") if the argument is valid, or enter | (for "incorrect") if any part of the argument is flawed. (Note: if the conclusion is true but the argument that led to it was wrong, you must enter I.) ☐ 1. For all n > 1, seriesΣ In(n) In(n) converges. 2, 1, arctan(n) the series arctan(n) n³ ☐ 4. For all n > 1, 123 converges. 1 n ln(n) series In(n) diverges. 2n . and the seriesΣconverges, so by the Comparison Test, 2, 3, and the series converges, so by the Comparison Test, the series-3 1 converges. ☐ 6. For all n > 2, In(n) >, and the series Σ converges, so by the Comparison Test, the seriesΣ In(n) converges.

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Chapter 3 Solutions

APPLIED CALCULUS FOR MGRL, LIFE, SOC SCI

Ch. 3.1 - State the following rules of differentiation in...Ch. 3.1 - Prob. 2CQ Ch. 3.1 - Prob. 3CQ Ch. 3.1 - Prob. 4CQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E

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Chapter 3 Solutions