Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 1 - Functions and Their Graphs - Section 1.1 Functions - 1.1 Assess Your Understanding - Page 55: 87

Answer

$A=8$ $f$ is not defined at $x=3$.

Work Step by Step

It is given, that $f(4)=0$. By substituting $x=4$ into the function $f(x)$ we get: $f(4)=\dfrac{2(4)-A}{2-3}$ $0=\dfrac{8-A}{-1}$ Solving this equation for $A$ we get: $\frac{8-A}{-1}=0$ $8-A=0(-1)$ $8-A=0\\ 8-A+A=0+A \\8=A$ $f$ is not defined when the denominator equals $0$. In this case it occurs, when: $x-3=0$ $x=3$, meaning that $f(3)$ is not defined.
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