University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 13: 67



Work Step by Step

$\mathrm{See\:the\:graph\:above.}$ $a.\:\:\:$ By observing the graph, we come to know that the point of intersections are $\:(-2,-1)\:$ and $\:(4,2).$ Based on the graph, the function $\:y=\frac{x}{2}\:$ (represented by thick black line) is above the function $\:y=1+\frac{4}{x}\:$ (represented by red thick curve), when: $-2 < x < 0\:\:$ or $\:\:x>4$ $b.\:\:\:$ $\frac{x}{2}>1+\frac{4}{x}$ $\frac{x}{2}-\frac{4}{x}-1>0$ $\frac{x^2-8-2x}{2x}>0$ Fraction is positive if both numerator and denominator are positive. $\mathrm{Case\:I:}$ $x^2-8-2x > 0,\:$ zeros of the function are -2 and 4, and the leading coefficient is positive. So, we got a parabola opening upwards, meaning that it is $\:> 0\:$ on $(\:-\infty,-2)\:$ and $(\:4,\infty).$ $2x> 0\:$ when $\:a>0$ We need both to be satisfied, so in the first case we have $\:x\in(4,\infty).$ $\mathrm{Case\:II:}$ $x^2-8-2x < 0\:$ when $(-2,4)$ $2x < 0\:$ when $\:x\in(-\infty,0)$ Solutions of this case are on $\:(-2,0)$ Combined solution is $\:(-2,0)\cup(4,\infty)$
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