## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 68

#### Answer

$$g\left( x \right) = {x^7} - {x^4} + 12x + 12$$

#### Work Step by Step

\eqalign{ & g'\left( x \right) = 7{x^6} - 4{x^3} + 12 \cr & g\left( x \right) = \int {g'\left( x \right)} dx \cr & then \cr & g\left( x \right) = \int {\left( {7{x^6} - 4{x^3} + 12} \right)} dx \cr & find{\text{ the general solution}} \cr & g\left( x \right) = {x^7} - {x^4} + 12x + C \cr & {\text{using the initial condition }}g\left( 1 \right) = 24 \cr & 24 = {\left( 1 \right)^7} - {\left( 1 \right)^4} + 12\left( 1 \right) + C \cr & 24 = 1 - 1 + 12 + C \cr & C = 12 \cr & {\text{the solution to the initial value problem is}} \cr & g\left( x \right) = {x^7} - {x^4} + 12x + 12 \cr}

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