Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 162: 5

Answer

$$ y=x-1, f'(1)=1$$

Work Step by Step

Given $$f(x)=x^{2}-x, \quad a=1$$ Since $f(1)=0$ and $$ f^{\prime}(a) =\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}$$ Then \begin{aligned} f^{\prime}(1) & =\lim _{h \rightarrow 0} \frac{(1+h)^{2}-(1+h)-\left(0\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1+2 h+h^{2}-1-h}{h}\\ &=\lim _{h \rightarrow 0}(1+h)=1 \end{aligned} Then the tangent line at $x=1$ is given by \begin{align*} \frac{y-y_1}{x-x_1}&=f'(a)\\ \frac{y}{x-1}&=1\\ \end{align*} Hence $$ y=x-1$$
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