Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 26

Answer

$$\left. {\bf{a}} \right)y = \frac{1}{{11}}x + \frac{{30}}{{11}},\,\,\,\,\,\,\,\left. {\bf{b}} \right)y = \frac{1}{{10}}x + \frac{5}{2}$$

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sqrt x ;\,\,\,\,\,\,\,\,\,{x_1} = 25,\,\,\,\,\,\,{x_2} = 36 \cr & \cr & \left. {\bf{a}} \right) \cr & {\text{Evaluate }}f\left( {{x_1}} \right){\text{ and }}f\left( {{x_2}} \right) \cr & {y_1} = f\left( {{x_1}} \right) = \sqrt {25} = 5 \cr & {y_1} = 5 \cr & \left( {{x_1},{y_1}} \right) = \left( {25,5} \right) \cr & and \cr & {y_2} = f\left( {{x_2}} \right) = f\left( {36} \right) = \sqrt {36} = 6 \cr & {y_2} = 6 \cr & \left( {{x_2},{y_2}} \right) = \left( {36,6} \right) \cr & \cr & {\text{The slope of the secant line is }} \cr & m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \frac{{6 - 5}}{{36 - 25}} = \frac{1}{{11}} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 5 = \frac{1}{{11}}\left( {x - 25} \right) \cr & y - 5 = \frac{1}{{11}}x - \frac{{25}}{{11}} \cr & y = \frac{1}{{11}}x + \frac{{30}}{{11}} \cr & \cr & \left. {\bf{b}} \right) \cr & {\text{Using the definition of the derivative}}: \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {x + h} - \sqrt x }}{h}\left( {\frac{{\sqrt {x + h} + \sqrt x }}{{\sqrt {x + h} + \sqrt x }}} \right) \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {\sqrt {x + h} } \right)}^2} - {{\left( {\sqrt x } \right)}^2}}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{{x + h - x}}{{h\left( {\sqrt {x + h} + \sqrt x } \right)}} \cr & f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\sqrt {x + h} + \sqrt x }} \cr & f'\left( x \right) = \frac{1}{{2\sqrt x }} \cr & \cr & {\text{The slope of the tangent line at the point }}\left( {{x_1},{y_1}} \right) = \left( {25,5} \right){\text{ is}} \cr & f'\left( {25} \right) = \frac{1}{{2\sqrt {25} }} \cr & f'\left( {25} \right) = \frac{1}{{10}} \cr & \cr & {\text{Find the equation of the line by the formula }}y - {y_1} = m\left( {x - {x_1}} \right) \cr & y - 5 = \frac{1}{{10}}\left( {x - 25} \right) \cr & y - 5 = \frac{1}{{10}}x - \frac{5}{2} \cr & y = \frac{1}{{10}}x + \frac{5}{2} \cr} $$
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