Problem 63 - Essential Calculus

Question

Let \(0

Step-by-Step Solution

Answer

When applying the Mean Value Theorem on the continuous and differentiable function \(f(x) = \sqrt{x}\) on the interval [a,b], it can be proven that \(\sqrt{b}-\sqrt{a}<\frac{b-a}{2 \sqrt{a}}\). The key was to observe that \(c>a\) and hence, \(\sqrt{c}>\sqrt{a}\). Therefore, \(\frac{1}{2\sqrt{c}}<\frac{1}{2\sqrt{a}}\) which in turn led to the required result.

Step 1: Define the function

Choose the function \(f(x) = \sqrt{x}\), because we have two square roots in the inequality. This function is continuous on the interval [a, b] and differentiable on the open interval (a, b), therefore the conditions of the Mean Value Theorem apply.

Step 2: Apply the Mean Value Theorem

The Mean Value Theorem states that there exists some \(c \in (a, b)\) such that \(f'(c)= \frac{f(b)-f(a)}{b-a}\). If you compute this, you find \(f'(c) = \frac{\sqrt{b} - \sqrt{a}}{b-a}\). Therefore, there must be some \(c \in (a, b)\) such that this is true.

Step 3: Find \(f'(c)\)

We find \(f'(x) = \frac{1}{2\sqrt{x}}\) by the derivative rule. Substitute \(x = c\) into this equation, we get \(f'(c) = \frac{1}{2\sqrt{c}}\).

Step 4: Relate \(f'(c)\) to \(\frac{\sqrt{b}-\sqrt{a}}{b - a}\)

Therefore, \(\frac{\sqrt{b}-\sqrt{a}}{b - a} = \frac{1}{2\sqrt{c}}\). As \(c \in (a, b)\), we have \(c > a\), that simply means \(\sqrt{c} > \sqrt{a}\) and hence, \(\frac{1}{2\sqrt{c}} < \frac{1}{2\sqrt{a}}\). This means \(\frac{\sqrt{b}-\sqrt{a}}{b - a} < \frac{1}{2\sqrt{a}}\) which rearranges to \(\sqrt{b}-\sqrt{a} < \frac{b - a}{2\sqrt{a}}\)

Key Concepts

Differentiable Functions

Understanding differentiable functions is crucial for learning the Mean Value Theorem (MVT). A function is differentiable on an interval if its derivative exists at every point in that interval. In simple terms, for differentiable functions, you can draw a tangent line at every point.
For the exercise, we use the function \( f(x) = \sqrt{x} \). This function is continuous and differentiable on \((a, b)\) because you can calculate a derivative for values between any two positive numbers \(a\) and \(b\) without any break or sharp corner in its graph.

• Continuity: The function must not have any gaps between \(a\) and \(b\).
• Differentiability: The function should have a smooth curve, allowing a tangent line.

Using the Mean Value Theorem is possible here because \( f(x) = \sqrt{x} \) fits these criteria.

Inequality Proofs

Inequality proofs, like the one in the exercise, involve showing that one side of an equation is less than or greater than the other. Here, we want to prove \( \sqrt{b} - \sqrt{a} < \frac{b-a}{2 \sqrt{a}} \).
The approach involves using the Mean Value Theorem, which gives a specific point \(c\) in \( (a, b) \), making the derivative equal to the slope of the line through \(f(b)\) and \(f(a)\).
By substituting \( f'(c) = \frac{\sqrt{b} - \sqrt{a}}{b-a} \) and evaluating this at \(c\), the inequality becomes clear as long as \( \sqrt{c} > \sqrt{a} \) (since \( c > a \)). This step transforms it into something we can compare, allowing the proof to logically follow from the derived equations.
In essence, inequalities often involve comparing values at certain key points as demonstrated through logical substitutions and transformations.

Derivative Rules

Derivative rules are tools used to find the slope of a function's curve. They tell us how the function's value changes with a tiny change in the input value.
For the function \( f(x) = \sqrt{x} \), the derivative is found using the power rule. Since \( \sqrt{x} = x^{1/2} \), we apply the rule that says the derivative of \( x^n \) is \( nx^{n-1} \). This gives us \( f'(x) = \frac{1}{2\sqrt{x}} \).

• Power Rule: A basic rule that calculates the derivative of \( x^n \).
• Substitution: Substitute \( x = c \) to find \( f'(c) \).

The derivative tells us how quickly \( \sqrt{x} \) changes, which the MVT uses to find a horizontal slope equaling \( \frac{\sqrt{b} - \sqrt{a}}{b-a} \). Understanding how to compute derivatives gives you the ability to solve complex problems like this one with ease.