Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

Show that \(\sqrt {1 + x} < 1 + \frac{1}{2}x\) if \(x > 0\).

For, \(x > 0,\sqrt {1 + x} < 1 + \frac{1}{2}x\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

The given inequality is \(\sqrt {1 + x} < 1 + \frac{1}{2}x\).

Step 2: Concept of Mean value theorem

“Let \(f\) be a function that satisfies the following hypothesis:

1. \(f\) is continuous on the closed interval \(\left( {a\;,{\rm{ }}b} \right)\).

2. \(f\) is differentiable on the open interval \(\left( {a,\;b} \right)\).

Then, there is a number \(c\) in \(\left( {a\;,{\rm{ }}b} \right)\) such that \({f^\prime }(c) = \frac{{f(b) - f(a)}}{{b - a}}\).

Or, equivalently, \(f(b) - f(a) = {f^\prime }(c)(b - a)\).”

Step 3: Calculation to show the give inequality

The equation \(y = 1 + \frac{1}{2}x\) is a line with slope 1 and \(y\) intercept 2 and let, \(f(x) = \sqrt {1 + x} \).

Apply Mean Value theorem with \(f(x) = \sqrt {1 + x} \) on the interval, \((0,\;a)\).

\(\begin{aligned}{c}\frac{{f(b) - f(a)}}{{b - a}} &= {f^\prime }(c)\\\frac{{\sqrt {a + 1} - \sqrt {0 + 1} }}{{a - 0}} &= \frac{1}{{2\sqrt {1 + c} }}\\\frac{{\sqrt {a + 1} - 1}}{a} &= \frac{1}{{2\sqrt {1 + c} }}\end{aligned}\)

It is known that for \(c > 0,\;\frac{1}{{2\sqrt {1 + c} }} < \frac{1}{2}\).

Therefore, the equation \(\frac{{\sqrt {a + 1} - 1}}{a} = \frac{1}{{2\sqrt {1 + c} }}\) becomes:

\(\begin{aligned}{c}\frac{{\sqrt {a + 1} - 1}}{a} < \frac{1}{2}\\\sqrt {a + 1} - 1 < \frac{1}{2}a\end{aligned}\)

Add \(1\) on both sides of the above equation.

\(\sqrt {a + 1} < \frac{1}{2}a + 1\)

Replace \(a\) by \(x\).

\(\sqrt {x + 1} < \frac{1}{2}x + 1\)

Thus, for \(x > 0,\sqrt {1 + x} < 1 + \frac{1}{2}x\).

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Show that \(\sqrt {1 + x} < 1 + \frac{1}{2}x\) if \(x > 0\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to show the give inequality

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Essential Calculus: Early Transcendentals

Answers without the blur.

Short Answer

Show that \(\sqrt {1 + x} < 1 + \frac{1}{2}x\) if \(x > 0\).

Step by Step Solution

Step 1: Given data

Step 2: Concept of Mean value theorem

Step 3: Calculation to show the give inequality

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