Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 2 - Section 2.1 - Functions - Exercises: 24

Answer

Evaluate $h(t) = t + \frac{1}{t}$: $h(-1) = -2$ $h(2) =\frac{5}{2} = 2.5 $ $h(\frac{1}{2}) =\frac{5}{2} = 2.5 $ $h(x-1) = (x-1) + \frac{1}{x-1} = \frac{x^2-2x+2}{x-1}$ $h(\frac{1}{x}) = \frac{1}{x} + x = \frac{x^{2}+1}{x}$

Work Step by Step

for $t = -1$ $h(-1) = (-1) + \frac{1}{-1}$... simplify fraction $= (-1) + (-1)$... add $=-2 $ _____________________ for $t = 2$ $h(2) = (2) + \frac{1}{2}$... convert whole number to fraction with common denominator $= \frac {4}{2} + \frac{1}{2}$... add $=\frac{5}{2} = 2.5 $ _____________________ for $t = \frac{1}{2}$ $h(\frac{1}{2}) = \frac{1}{2} + \frac{1}{\frac{1}{2}}$... simplify complex fraction $= \frac{1}{2} + 2$... convert whole number to fraction with common denominator $=\frac{1}{2} + \frac{4}{2} $ ...add $=\frac{5}{2}=2.5$ _____________________ for $t = x-1$ $h(x-1) = (x-1) + \frac{1}{x-1}$... convert polynomial to fraction with common denominator $= \frac{(x-1)^2}{x-1} + \frac{1}{x-1}$... add $= \frac{(x-1)^2+1}{x-1}$ ... square (x-1) and add 1 $= \frac{x^2-2x+2}{x-1}$ _____________________ for $t = \frac{1}{x}$ $h(\frac{1}{x}) = \frac{1}{x} + \frac{1}{\frac{1}{x}}$... simplify complex fraction $= \frac{1}{x} + x$... convert variable to fraction with common denominator $=\frac{1}{x} + \frac{x^{2}}{x} $ ...add $=\frac{x^{2}+1}{x}$ _____________________
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