Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 116: 85



Work Step by Step

Since $$ f'(x)= 2x + 3$$ Then at $x=2, $ $m= f'(2)=7 $. Hence, the tangent line is \begin{align*} \frac{y-y_1}{x-x_1}&=m\\ \frac{y-10}{x-2}&=7\\ y&=7x-4 \end{align*} Since the tangent line intersect with the $x-$ axis at $x=4/7$, then $Q $ has coordinates $( 4/ 7, 0)$, $R $ has coordinates $(2, 0)$, and the subtangent is $$ 2-\frac{4}{7}=\frac{10}{7}$$
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