Problem 24 - Precalculus Mathematics for Calculus

Question

Find \(f^{\prime}(a),\) where \(a\) is in the domain of \(f .\) $$f(x)=\sqrt{x-2}$$

Step-by-Step Solution

Answer

The derivative \( f'(a) \) is \( \frac{1}{2\sqrt{a-2}} \), where \( a \geq 2 \).

Step 1: Understand the Problem

We need to find the derivative of the function \( f(x) = \sqrt{x-2} \) at a point \( x = a \). This means we are looking for \( f'(a) \).

Step 2: Determine the Derivative Function

The function \( f(x) = \sqrt{x-2} \) can be rewritten as \( f(x) = (x-2)^{1/2} \). To find the derivative, use the power rule: \( \frac{d}{dx} (g(x))^n = n(g(x))^{n-1}g'(x) \). Here, \( n = \frac{1}{2} \) and \( g(x) = x-2 \), so \( g'(x) = 1 \).

Step 3: Apply the Power Rule

Using the power rule, the derivative is \( f'(x) = \frac{1}{2}(x-2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-2}} \).

Step 4: Evaluate the Derivative at \( x=a \)

Now that we have \( f'(x) = \frac{1}{2\sqrt{x-2}} \), substitute \( x = a \) into this equation to find \( f'(a) \). The result is \( f'(a) = \frac{1}{2\sqrt{a-2}} \).

Step 5: Determine the Domain

The domain of \( f(x) = \sqrt{x-2} \) is \( x \geq 2 \), because the expression under the square root must be non-negative. Therefore, \( a \) must also satisfy \( a \geq 2 \) for \( f'(a) \) to be defined.

Key Concepts

Power Rule

The power rule is a simple yet powerful tool in calculus used to differentiate functions with ease. It is especially useful for functions that can be written as a power of another expression.
The generalized formula for the power rule is:

• If you have a function of the form \( g(x) = (h(x))^n \), the derivative is \( g'(x) = n(h(x))^{n-1}h'(x) \).

To apply it, follow these steps:- Identify the power \( n \) of your function.- Differentiate the inner function, \( h(x) \), and call it \( h'(x) \).- Apply the power rule formula by multiplying \( n \) by the power of \( h(x) \) decreased by one, and multiplying again by the derivative of the inner function, \( h'(x) \).
In our exercise, the function \( f(x) = (x-2)^{1/2} \) fits perfectly into this rule with \( n = \frac{1}{2} \). The inner function is \( x-2 \), which has a derivative of 1. So, applying the power rule gives us:\[ f'(x) = \frac{1}{2}(x-2)^{-1/2} \cdot 1 \]resulting in the derivative \( \frac{1}{2\sqrt{x-2}} \).

Square Root Function

A square root function involves the square root of an expression, such as \( \sqrt{x} \). These functions have special characteristics and require particular attention when differentiating.
For a typical square root function, \( f(x) = \sqrt{x} \), it can be rewritten using exponents as \( f(x) = x^{1/2} \). This transformation is crucial for applying derivative rules like the power rule.
In our problem, the square root function \( \sqrt{x-2} \) is transformed as \((x-2)^{1/2}\), allowing us to apply the power rule seamlessly.
Understanding this conversion from radical to exponent form simplifies differentiation and helps in visualizing how the function behaves.
Square root functions are restricted to non-negative domains since square roots of negative numbers are not real within the real number system. This restriction affects the domain, which we will discuss further in the next section.

Domain of a Function

The domain of a function is the set of all possible input values (or \( x \)-values) for which the function is defined.
For functions involving roots, such as square roots, the expression under the root sign (the radicand) must be non-negative to ensure the output is a real number.

• In our function \( f(x) = \sqrt{x-2} \), the radicand is \( x-2 \).
• This means the inequality \( x-2 \geq 0 \) must hold true.
• Solving this inequality results in \( x \geq 2 \).

Therefore, the domain of \( f(x) = \sqrt{x-2} \) is \( x \geq 2 \). This condition ensures that both the function and its derivative, \( f'(x) = \frac{1}{2\sqrt{x-2}} \), are well-defined and yields real values. In practical terms, it indicates that you can only plug in values of \( x \) greater than or equal to 2. Keeping track of the domain is crucial, as it determines the valid inputs for function evaluation and differentiation, ensuring mathematical accuracy and relevance.