Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.4 - Rational Expressions - 1.4 Exercises - Page 45: 103

Answer

(a) Smallest possible value for this expression is $2$. (b)$x+\frac{1}{x}\geq 2$ $x^2 + \frac{x}{x} \geq 2x$ $x^2-2x +1\geq 0$ $x∈(0, +\infty)$

Work Step by Step

(a) Let's put the values provided and consider the value of expression: $x=1; x+\frac{1}{x} = 1 + \frac{1}{1} = 2$ $x=3; x+\frac{1}{x} = 3 +\frac{1}{3} = 3.(3)$ $x=\frac{1}{2}; x+\frac{1}{x} = \frac{1}{2}+\frac{2}{1} = 2.5$ $x=\frac{9}{10}; x+\frac{1}{x} = \frac{9}{10} + \frac{10}{9} = 2.0(1)$ $x=\frac{99}{100}; x+\frac{1}{x} = \frac{99}{100} + \frac{100}{99} = 2.000(10)$ I will input two more value for a better visualization of output from this expression: $x=1.001; x+\frac{1}{x} = 1.001+\frac{1}{1.001} = 2.000,000(999,000)$ $x=0.0001; x+\frac{1}{x} = 0.0001+\frac{1}{0.0001} = 10,000.0001$ As you can see the value of this expression varies from $2$ (as $x$ approaches $1$) to $+\infty$ (as $x$ approaches $0$ or $+\infty$). So the minimum value is $2$ when $x=1$. (b) Let's try to simplify it: $x+\frac{1}{x}\geq 2$ //Multiply by $x$ $x^2 + \frac{x}{x} \geq 2x$ //Move terms to one side $x^2-2x +1\geq 0$ //Solve for quadratic inequality $x=\frac{-b±\sqrt{b^2-4ac}}{2a} = \frac{2+\sqrt{4-4}}{2}=1$ We have: $x∈(-\infty, 0)U(0, +\infty)$, Buy we have restriction $x>0$, so the domain of definition is $x∈(0, +\infty)$ So, for any non-negative $x$ value we get output greater or equal to $2$. We have also proven it with actual numbers in (a)
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