Problem 15 - Calculus: Early Transcendentals

Question

Show that \(\sqrt{1+x} \leq 1+(x / 2)\) if \(-1

Step-by-Step Solution

Answer

The inequality holds for \(-1 < x < 1\) because \( \sqrt{1+x} \leq 1 + \frac{x}{2} \).

Step 1: Define the Function

Let us define the function \( f(x) = \sqrt{1 + x} - \left(1 + \frac{x}{2}\right) \). We need to show that \( f(x) \leq 0 \) for \(-1 < x < 1\).

Step 2: Compute Derivative

Compute the derivative \( f'(x) \) of \( f(x) \). The derivative of \( f(x) = \sqrt{1 + x} - 1 - \frac{x}{2}\) is \[ f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2}. \]

Step 3: Analyze the Derivative

Examine \( f'(x) \): For \(-1 < x < 1\), \( \sqrt{1 + x} \) is positive, hence \( f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2} \leq 0 \).This inequality holds because \( \sqrt{1+x} \geq 1 \) for \(x \geq 0\), causing \( f'(x) \) to be non-positive.

Step 4: Confirm Function Behavior

Since \( f(x) \leq 0 \) when \(x = 0\) (as \( f(0) = 0 \)), and \( f'(x) \leq 0 \) for \(-1 < x < 1\), \( f(x) \) is non-increasing in this interval.This means \( f(x) \leq 0 \) for all \(-1 < x < 1\), proving \( \sqrt{1+x} \leq 1 + \frac{x}{2} \).

Step 5: Check Boundary Conditions

Evaluate the boundary conditions: For \(x = 0\), both sides are equal, confirming \( \sqrt{1+0} = 1 + \frac{0}{2} \).As \(x\) approaches \(-1\) or \(1\), the inequality holds since \( \sqrt{1+x} \leq 1 + \frac{x}{2} \).

Key Concepts

Derivative Analysis

In calculus, analyzing the derivative of a function is key to understanding how the function behaves over specific intervals. Let's look at the function given by the exercise:
\[ f(x) = \sqrt{1+x} - \left( 1 + \frac{x}{2} \right). \]
To determine whether this function is always non-positive for \(-1 < x < 1\), we start by finding its derivative. The derivative of a function provides insight into its rate of change.
For our function, the derivative is computed as:
\[ f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2}. \]
This expression shows how the function changes with respect to x.
Through derivative analysis, we can determine whether the function is increasing or decreasing. It allows us to explore the function's behavior in more depth and helps validate the inequality.

Function Behavior

Understanding the behavior of a function involves observing how it changes across a given interval. Function behavior is closely tied to its derivative, as the sign of the derivative indicates whether the function is increasing, decreasing, or remaining constant.
For the function \( f(x) = \sqrt{1+x} - \left( 1 + \frac{x}{2} \right) \), examining its derivative \( f'(x) = \frac{1}{2\sqrt{1+x}} - \frac{1}{2} \) reveals significant insights:

• For values \(x\) such that \(-1 < x < 1\), the derivative is always non-positive, signifying that the function is non-increasing in this range.

This means that as \(x\) moves through this interval, \( f(x)\) either remains constant or decreases, which is crucial to verifying the inequality in the exercise.

Boundary Conditions

Checking boundary conditions is essential for assessing how a function behaves at the endpoints of an interval.
For the function \( f(x) = \sqrt{1+x} - \left(1 + \frac{x}{2}\right) \), the interval of interest is \(-1 < x < 1\).
Here's what to consider at the boundaries:

• At \(x = 0\), \( f(0) = 0 \), demonstrating that both sides of the original inequality are equal.
• As \(x\) approaches \(-1\) or \(1\), the inequality \(\sqrt{1+x} \leq 1 + \frac{x}{2}\) remains valid, further confirming the constant or decreasing behavior of the function.

These boundary checks ensure the behavior of the function remains consistent and supports the claim made in the exercise.

Non-Increasing Functions

A non-increasing function is one that either remains constant or decreases over its entire domain.
In this exercise, we established that the derivative of the function \( f(x) = \sqrt{1+x} - \left(1 + \frac{x}{2}\right) \) is non-positive for \(-1 < x < 1\).
This indicates \( f(x) \) does not increase in this interval.
Therefore:

• For any point in the specified domain, \( f(x) \) can only maintain or decrease its value, never surpassing 0.
• Ultimately, this behavior corroborates the inequality \( \sqrt{1+x} \leq 1 + \frac{x}{2} \), proving it to be true within the designated range.

Understanding the concept of non-increasing functions is pivotal in verifying inequalities such as the one featured in our exercise. It showcases the applicability of derivative signs in determining overall function behavior.