Problem 29 - Calculus Early Transcendentals

Question

(a) Use the Mean-Value Theorem to show that $$ \sqrt{y}-\sqrt{x}<\frac{y-x}{2 \sqrt{x}} $$ if \(0

Step-by-Step Solution

Answer

Part (a) uses the Mean Value Theorem; Part (b) applies it to prove the desired inequality.

Step 1: Understanding the Problem

We are using the Mean Value Theorem to show inequalities involving square roots. We need to prove part (a) first about the difference of square roots and then apply it to verify part (b).

Step 1: Applying Mean Value Theorem for Part (a)

We consider the function \( f(t) = \sqrt{t} \). The Mean Value Theorem states that there exists a point \( c \in (x, y) \) such that \( f'(c) = \frac{f(y) - f(x)}{y - x} \). Calculate \( f'(t) = \frac{1}{2\sqrt{t}} \).

Step 2: Estimating the Derivative

Since \( c \in (x, y) \), \( x < c < y \), it follows that \( f'(c) = \frac{1}{2\sqrt{c}} < \frac{1}{2\sqrt{x}} \). This gives \[ \frac{\sqrt{y} - \sqrt{x}}{y-x} < \frac{1}{2\sqrt{x}} \].

Step 3: Rearranging for Part (a)

Multiplying both sides by \( y-x \), we obtain \( \sqrt{y} - \sqrt{x} < \frac{y-x}{2\sqrt{x}} \), which was to be shown for part (a).

Step 4: Using Result of Part (a) for Part (b)

Rearrange the inequality in part (a) to \( 2\sqrt{x} (\sqrt{y} - \sqrt{x}) < y - x \). Simplifying gives \( 2 \sqrt{xy} < x + y \), which states \( \sqrt{xy} < \frac{x + y}{2} \).

Step 5: Conclusion of Part (b)

The inequality from part (a) implies the required result in part (b), proving that \( \sqrt{x y} < \frac{1}{2}(x+y) \) for \( 0 < x < y \).

Key Concepts

Inequalities Involving Square Roots

Let's delve into the concept of inequalities involving square roots, which plays a vital role in understanding advanced mathematical concepts. Inequalities with square roots often appear in calculus and algebra, where they are used to compare quantities that are not directly relatable through addition or subtraction alone. In these cases, revealing the relationship between quantities involves using functions like the square root, which can sometimes complicate inequalities slightly due to their non-linear nature.

Consider an inequality where you're tasked to show that a manipulated square root expression, like \(\sqrt{y} - \sqrt{x}\), is less than another expression. Here, the trick is to utilize properties specific to square roots:

• Square roots are always positive for non-negative numbers.
• The difference \(\sqrt{y} - \sqrt{x}\) is meaningful only if \(y > x\), since square roots preserve the order of numbers.

When using calculus-based techniques, such as the Mean Value Theorem, inequalities involving square roots can efficiently be handled. Such theorems help us find intermediate values that can simplify comparisons, turning seemingly complex expressions into more manageable forms.

Calculus

Calculus is the branch of mathematics that deals with change and motion, using derivatives and integrals as its primary tools. It provides powerful methods to analyze functions and solve a variety of problems by examining the rate at which things change. The Mean Value Theorem, a fundamental part of calculus, assists us in discovering how a function behaves over a particular interval.

The Mean Value Theorem (MVT) states that for any differentiable function \(f\) on an open interval \((a, b)\), there exists a point \(c\) in \((a, b)\) such that:

• \(f'(c) = \frac{f(b) - f(a)}{b - a}\)

This theorem is crucial for establishing inequalities and understanding the behavior of functions. It essentially provides a bridge between the average rate of change over an interval and the instantaneous rate of change at a specific point.

When applying the MVT to a function like \(f(t) = \sqrt{t}\), it helps to determine conditions like those found in inequalities involving square roots. This operation simplifies the analysis of how function values relate to each other when differentiating and integrating under particular constraints.

Derivatives

In calculus, derivatives are essential as they measure how a function changes as its input changes. They are vital for applications ranging from physics to engineering and even economics, where the rate of change is significant. The concept of a derivative is foundational for both understanding and utilizing calculus.

To calculate a derivative, one must understand that it represents the slope of a function at a given point. It is mathematically defined as:

• \(f'(t) = \lim_{{h \to 0}} \frac{f(t + h) - f(t)}{h}\)

In our task of using the Mean Value Theorem to deal with inequalities involving square roots, the derivative of \(f(t) = \sqrt{t}\) plays a key role. The derivative, \(f'(t) = \frac{1}{2\sqrt{t}}\), helps find the exact point where certain conditions set by the inequality hold true.

Derivatives provide insights into the mechanics behind this expression and allow us to explore deeper into the properties of the function \(f(t) = \sqrt{t}\) over an interval. They allow the translation of complex functional relationships into simpler mathematical comparisons, which can be crucial for verifying inequalities or finding optimal solutions in mathematical problems.