Problem 27 - Calculus: Early Transcendentals

Question

Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative. \( g(x) = \sqrt{9 - x} \)

Step-by-Step Solution

Answer

The derivative is \( g'(x) = \frac{-1}{2\sqrt{9-x}} \). The domain of \( g(x) \) is \( x \leq 9 \) and of \( g'(x) \) is \( x < 9 \).

Step 1: Understand the Function

The function provided is \( g(x) = \sqrt{9 - x} \). This is a square root function which implies that the expression inside the root, \( 9 - x \), must be greater than or equal to zero. Consequently, the domain of \( g(x) \) is \( x \leq 9 \).

Step 2: Determine the Difference Quotient

The definition of the derivative is given by the limit \( \lim_{h \to 0} \frac{g(x+h) - g(x)}{h} \). For our function: \[ g'(x) = \lim_{h \to 0} \frac{\sqrt{9 - (x+h)} - \sqrt{9 - x}}{h} \]

Step 3: Rationalize the Numerator

To simplify the difference of square roots, multiply and divide by the conjugate: \( \lim_{h \to 0} \frac{(\sqrt{9-x-h} - \sqrt{9-x})(\sqrt{9-x-h} + \sqrt{9-x})}{h(\sqrt{9-x-h} + \sqrt{9-x})} \). This yields: \[ \lim_{h \to 0} \frac{(9 - x - h) - (9 - x)}{h(\sqrt{9-x-h} + \sqrt{9-x})} \] which simplifies to \[ \lim_{h \to 0} \frac{-h}{h(\sqrt{9-x-h} + \sqrt{9-x})} \].

Step 4: Simplify the Expression

Cancel \( h \) in the numerator and the denominator: \[ \lim_{h \to 0} \frac{-1}{\sqrt{9-x-h} + \sqrt{9-x}} \].

Step 5: Apply the Limit

As \( h \to 0 \), \( \sqrt{9-x-h} \to \sqrt{9-x} \). Thus, the derivative is: \[ g'(x) = \lim_{h \to 0} \frac{-1}{\sqrt{9-x} + \sqrt{9-x}} = \frac{-1}{2\sqrt{9-x}} \].

Step 6: Determine the Domain of the Derivative

The domain of the derivative \( g'(x) = \frac{-1}{2\sqrt{9-x}} \) is the set where \( 9 - x \) is greater than zero to avoid division by zero. Hence, the domain is \( x<9 \).

Key Concepts

Domain of a Function

The domain of a function is essentially the set of all possible input values (typically known as x-values) for which the function is defined. If a function includes a square root, such as in the function \( g(x) = \sqrt{9 - x} \), we need the expression under the square root to be non-negative, since you cannot take the square root of a negative number in the real number system.

• For \( g(x) = \sqrt{9 - x} \), the condition 9 - x \( \geq 0 \) must be satisfied.
• Solving this inequality yields \( x \leq 9 \).

Thus, the domain of the function \( g(x) \) is all real numbers less than or equal to nine, or \( x \leq 9 \). This ensures the output is a real number.

Rationalize the Numerator

Rationalizing the numerator is a technique used to simplify expressions involving square roots. It is particularly helpful when finding derivatives using limits, as it can eliminate complex radicals that make simplifying difficult. In the derivative process for \[ g'(x) = \lim_{h \to 0} \frac{\sqrt{9 - (x+h)} - \sqrt{9 - x}}{h} \], multiplying the numerator and the denominator by the conjugate pair is a key step. The conjugate for an expression \( a - b \) is \( a + b \).

• By applying \( \sqrt{9-x-h} + \sqrt{9-x} \) as the conjugate, the radicals in the numerator are eliminated.
• This transforms the expression into a simpler form: \[ \frac{(\sqrt{9-x-h})^2 - (\sqrt{9-x})^2}{h(\sqrt{9-x-h} + \sqrt{9-x})} \], allowing us to further reduce the expression.

By employing this method, you avoid complex arithmetic with roots, simplifying the process of finding the limit.

Limit Definition of Derivative

To find a derivative using the limit definition, you calculate the slope of the tangent line to a function at a given point. The formal definition is given by the limit \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \). For a function like \( g(x) = \sqrt{9 - x} \), this becomes \[ g'(x) = \lim_{h \to 0} \frac{\sqrt{9 - (x+h)} - \sqrt{9 - x}}{h} \].

• This requires evaluating the limit as \( h \) approaches zero.
• Through steps like rationalizing the numerator, you simplify the limit expression until the \( h \) in the denominator is canceled out, allowing you to find the exact slope at any point within the domain of \( x \).

This method is powerful for directly finding derivatives without needing derivative shortcuts.

Square Root Function

Square root functions, such as \( g(x) = \sqrt{9 - x} \), have their unique properties and domain restrictions, largely due to the nature of square roots. These functions only provide real outputs when the expression inside the root is non-negative. Consequently, the graph for \( g(x) \) begins at a converging point (x-value cutoff), maintaining values where the interior of the square root is zero or positive.

• In terms of behavior, as \( x \) approaches the endpoint of its domain from the left (i.e., just less than or equal to 9), the function values trend down, but the derivative provides insight into the rate of this change.
• Additionally, unlike polynomial functions, square root functions may not span across the entire set of real numbers due to their domain limitations.

Understanding square root functions in differentiation context emphasizes considering domain restrictions and leads smoothly into derivative calculations. These nuances are crucial for accurately analyzing and graphing such functions.