Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 922: 30

Answer

$f’\left( x\right) =\dfrac {1}{2\sqrt {x-2}}$ $f'\left( 3\right) =\dfrac {1}{2}$ $f'\left( 4\right) =\dfrac {1}{2\sqrt {2}}$

Work Step by Step

$f\left( x\right) =\sqrt {x-2}=\left( x-2\right) ^{\dfrac {1}{2}}\rightarrow f'\left( x\right) =\dfrac {1}{2}\times \left( x-2\right) ^{\dfrac {1}{2}-1}\times \left( x-2\right)’=\dfrac {1}{2}\times \left( x-2\right) ^{-\dfrac {1}{2}}\times \left( 1\times x^{1-1}-0\right) =\dfrac {1}{2\sqrt {x-2}}$ $f'\left( 3\right) =\dfrac {1}{2\sqrt {x-2}}=\dfrac {1}{2\times \sqrt {3-2}}=\dfrac {1}{2\times \sqrt {1}}=\dfrac {1}{2}$ $f'\left( 4\right) =\dfrac {1}{2\sqrt {x-2}}=\dfrac {1}{2\times \sqrt {4-2}}=\dfrac {1}{2\times \sqrt {2}}=\dfrac {1}{2\sqrt {2}}$
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