College Algebra 7th Edition

Published by Brooks Cole

Chapter 4, Exponential and Logarithmic Functions - Section 4.1 - Exponential Functions - 4.1 Exercises - Page 372: 9

Answer

$g(\frac{1}{2})\approx 0.192$; $g(\sqrt2) \approx 0.070$; $g(-3.5)\approx 15.588$; $g(\frac{1}{4}) \approx 1.552$

Work Step by Step

Evaluate the function for the given values of $x$ by substituting each $x$ into $g(x)$ then using a calculator to find the exact value. When x=$\frac{1}{2}$: $g(x) = \left(\dfrac{1}{3}\right)^{x+1} \\g(\frac{1}{2})=\left(\dfrac{1}{3}\right)^{\frac{1}{2}+1} \\g(\frac{1}{2})=\left(\dfrac{1}{3}\right)^{\frac{3}{2}} \\g(\frac{1}{2})\approx 0.192$ When $x=\sqrt2$: $g(x) = \left(\dfrac{1}{3}\right)^{x+1} \\g(\sqrt2) = \left(\dfrac{1}{3}\right)^{\sqrt2+1} \\g(\sqrt2) \approx 0.070$ When $x=-3.5$: $g(x)=\left(\dfrac{1}{3}\right)^{x+1} \\g(-3.5)=\left(\dfrac{1}{3}\right)^{-3.5+1} \\g(-3.5)=\left(\dfrac{1}{3}\right)^{-2.5} \\g(-3.5)\approx 15.588$ When $x=-1.4$: $g(x)=\left(\dfrac{1}{3}\right)^{x+1} \\g(\frac{1}{4}) = \left(\dfrac{1}{3}\right)^{-1.4+1} \\g(\frac{1}{4}) = \left(\dfrac{1}{3}\right)^{-0.4} \\g(\frac{1}{4}) \approx 1.552$

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