Explanation Given: The equation of the function is R = 75 − 18 i 2 + 8 i 3 − i 4 . Result used: The Second Derivative Test for Concavity: Let y = f ( x ) be twice differentiable on an interval I . 1. If f ″ > 0 on I , the graph of f over I is concave up. 2. If f ″ < 0 on I , the graph of f over I is concave down. Definition used: Critical point: A critical point of a function f is any number x in the domain of f such that f ′ ( x ) = 0 or f ′ ( x ) = 0 does not exist. Points of Inflection: A point ( c , f ( c ) ) where the graph of a function has a tangent line and where the concavity changes is a point of inflection. Calculation: Find the critical points by equating the first derivative to zero. 0 = 0 − 36 i + 24 i 2 − 4 i 3 − 9 i + 6 i 2 − i 3 = 0 i ( 9 − 6 i + i 2 ) = 0 i ( i − 3 ) 2 = 0 Simplify further to get, i ( i − 3 ) 2 = 0 i = 0 , 3 Choose the test value − 1 and 1 placed to the left and right of critical point t = 0 and choose the test value 1 and 4 placed to the left and right of critical point t = 3 . Substitute the test value i = − 1 in first derivative to check the sign of the derivative. R ′ ( − 1 ) = − 36 ( − 1 ) + 24 ( − 1 ) 2 − 4 ( − 1 ) 3 = 36 + 24 + 4 = 64 > 0 Substitute the test value i = 1 in first derivative to check the sign of the derivative. R ′ ( 1 ) = − 36 ( 1 ) + 24 ( 1 ) 2 − 4 ( 1 ) 3 = − 36 + 24 − 4 = − 16 < 0 Substitute the test value i = 4 in first derivative to check the sign of the derivative. R ′ ( 4 ) = − 36 ( 4 ) + 24 ( 4 ) 2 − 4 ( 4 ) 3 = − 144 + 384 − 256 = − 16 < 0 Thus, the function is decreasing for 0 < i < 3 and 3 < i < 5 and increasing for i < 0

11th Edition

ISBN: 9780134437705

Author: Washington

Publisher: PEARSON

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Chapter 24.5, Problem 49E

To determine

To Sketch: The graph of the function R=75−18i2+8i3−i4 if R≥0 and i can be positive and negative.

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Chapter 24.5, Problem 48E

Chapter 24.5, Problem 50E

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Chapter 24 Solutions

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The graph of the function R = 75 − 18 i 2 + 8 i 3 − i 4 if R ≥ 0 and i can be positive and negative.

To Sketch: The graph of the function R=75−18i2+8i3−i4 if R≥0 and i can be positive and negative.

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Chapter 24 Solutions