College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Section 3.1 - Functions - 3.1 Assess Your Understanding: 93



Work Step by Step

We are given: $f(x)=2x^{3}+\mathrm{A}x^{2}+4x-5$ and $f(2)=5$ We solve for $A$: $f(2)=2(2)^{3}+A(2)^{2}+4(2)-5$ $f(2)=16+4A+8-5$ $f(2)=4A+19$ Since $f(2)=5$: $5=4A+19$ $-14=4A$ $A=\frac{-14}{4}$ $A=-\frac{7}{2}$
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