University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 24

Answer

$f'(x) = 2x-3$

Work Step by Step

$f(x) = x^{2} - 3x +4$ $f'(x) = \lim\limits_{z \to x}\frac{(z^{2} - 3z +4) - x^{2} + 3x -4}{z-x}$ $f'(x) = \lim\limits_{z \to x}\frac{z^{2} - 3z - x^{2} + 3x}{z-x}$ $f'(x) = \lim\limits_{z \to x}\frac{(z+x)(z-x)-3(z-x)}{z-x}$ $f'(x) = \lim\limits_{z \to x}(z+x-3)$ $f'(x) = 2x-3$
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