Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises: 63

Answer

\[ = {x^2} + 2xa + {a^2} - 2\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = {x^3} - 2x \hfill \\ \hfill \\ \frac{{f\,\left( x \right) - f\,\left( a \right)}}{{x - a}} = \frac{{{x^3} - 2x - {a^3} + 2a}}{{x - a}} \hfill \\ \hfill \\ factor \hfill \\ \hfill \\ = \frac{{{x^3} - {a^3} - 2\,\left( {x - a} \right)}}{{x - a}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{\,\left( {x - a} \right)\,\left( {{x^2} + 2xa + {a^2}} \right) - 2\,\left( {x - a} \right)}}{{x - a}} \hfill \\ \hfill \\ = \frac{{\,\left( {x - a} \right)\,\left( {{x^2} + 2xa + {a^2} - 2} \right)}}{{x - a}} \hfill \\ \hfill \\ = {x^2} + 2xa + {a^2} - 2 \hfill \\ \end{gathered} \]
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