Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 3 - Section 3.7 - Polynomial and Rational Inequalities - 3.7 Exercises - Page 316: 38

Answer

$(0,1)$

Work Step by Step

We have to determine the values of $x$ for which $f(x) > g(x)$. Given $f(x) = \frac {1}{x}$ and $g(x) = \frac {1}{x-1}$, we have $\frac{1}{x} > \frac{1}{x-1}$ $\frac{1}{x} - \frac {1}{x-1} > 0 $ $\frac {-1}{(x)(x-1)} > 0$ $(x)(x-1) < 0$ Find the zeros of the expressions in the numerator AND the denominator $x = 0, x=1$ Test numbers in the intervals $(-\infty,0)$, $(0,1)$ and $(1,\infty)$to determine if the function is negative or positive: $(-\infty, 0)$: $\frac {(+)}{(-)(-)} = (+)$ $(0, 1)$: $\frac {(+)}{(+)(-)} = (-)$ $(1,\infty)$: $\frac {(+)}{(+)(+)} = (+)$ Thus the solution is $(0,1)$.
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