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Found in: Page 216

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# 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 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 $$x > 0,\sqrt {1 + x} < 1 + \frac{1}{2}x$$.