Explanation Formula used: The equation of the tangent line is, y − y 1 = m ( x − x 1 ) . Calculation: The given function is, y = − 3 x 5 − 8 x 3 + 4 x 2 (1) The slope (m) of the tangent line is computed as follows: Obtain the derivative y ' ( x ) of the function y = − 3 x 5 − 8 x 3 + 4 x 2 as follows. y ' ( x ) = d y d x = d d x [ − 3 x 5 − 8 x 3 + 4 x 2 ] = d d x [ − 3 x 5 ] + d d x [ − 8 x 3 ] + d d x [ 4 x 2 ] [ ∵ d d x ( u ( x ) + v ( x ) ) = d d x ( u ( x ) ) + d d x ( v ( x ) ) ] On further simplification, the derivative y ' ( x ) becomes, y ' ( x ) = − 3 [ d d x [ x 5 ] ] − 8 [ d d x [ x 3 ] ] + 4 [ d d x [ x 2 ] ] [ ∵ d d x ( k . g ( x ) ) = k . d d x ( g ( x ) ) ] = − 3 [ 5 x 5 − 1 ] − 8 [ 3 x 3 − 1 ] + 4 [ 2 x 2 − 1 ] [ ∵ d d x ( x n ) = n x n − 1 ] = − 15 x 4 − 24 x 2 + 8 x 1 That is, y ' ( x ) = − 15 x 4 − 24 x 2 + 8 x (2) Therefore, the derivative of the function y = − 3 x 5 − 8 x 3 + 4 x 2 is

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 12.1, Problem 32E

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To find: The slope and the equation of the tangent line of the function y=−3x5−8x3+4x2 at x=1 .

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Chapter 12.1, Problem 31E

Chapter 12.1, Problem 33E

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Chapter 12 Solutions

Finite Mathematics and Calculus with Applications (10th Edition)

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The slope and the equation of the tangent line of the function y = − 3 x 5 − 8 x 3 + 4 x 2 at x = 1 .

Solution Summary: The author explains that the slope and the equation of the tangent line are 31 and y=-31x+24.

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Chapter 12 Solutions