Problem 30 - Precalculus: Mathematics for Calculus

Question

Find the following for the given function \(f:\) (a) \(f^{\prime}(a),\) where \(a\) is in the domain of \(f,\) and (b) \(f^{\prime}(3)\) and \(f^{\prime}(4)\) $$f(x)=\sqrt{x-2}$$

Step-by-Step Solution

Answer

(a) \(f'(a)=\frac{1}{2\sqrt{a-2}}\) (b) \(f'(3) = \frac{1}{2}\), \(f'(4) = \frac{\sqrt{2}}{4}\).

Step 1: Find the Derivative f'(x)

The given function is \(f(x) = \sqrt{x-2}\). To find \(f'(x)\), we use the chain rule. Rewriting the function, we have \(f(x) = (x-2)^{1/2}\).Applying the chain rule: \(f'(x) = \frac{1}{2}(x-2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-2}}\). So, the derivative is \(f'(x) = \frac{1}{2\sqrt{x-2}}\).

Step 2: Determine f'(a) for a General x

Now that we have \(f'(x) = \frac{1}{2\sqrt{x-2}}\), we can express \(f'(a)\) for any \(a\) in the domain of the function.The function is defined for \(x \geq 2\), thus \(f'(a) = \frac{1}{2\sqrt{a-2}}\) when \(a \geq 2\).

Step 3: Calculate f'(3)

To find \(f'(3)\), substitute 3 into the expression for the derivative:\(f'(3) = \frac{1}{2\sqrt{3-2}} = \frac{1}{2\sqrt{1}} = \frac{1}{2}\).

Step 4: Calculate f'(4)

To find \(f'(4)\), substitute 4 into the expression for the derivative:\(f'(4) = \frac{1}{2\sqrt{4-2}} = \frac{1}{2\sqrt{2}} = \frac{1}{2 \cdot \sqrt{2}} = \frac{1}{2\sqrt{2}}\).To rationalize the denominator, multiply the fraction by \(\frac{\sqrt{2}}{\sqrt{2}}\), giving \(f'(4) = \frac{\sqrt{2}}{4}\).

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus that lets us find the derivative of a composite function. A composite function is when one function is nested inside another, like \(f(x) = \sqrt{x-2}\). Here, the inner function is \(g(x) = x-2\) and the outer function is \(h(u) = \sqrt{u}\), where \(u = g(x)\).
To apply the chain rule, follow these steps:

• First, find the derivative of the outer function \(h(u)\) with respect to \(u\), which is \(\frac{1}{2}u^{-1/2}\).
• Second, find the derivative of the inner function \(g(x)\) with respect to \(x\), which is \(1\).
• Lastly, multiply these derivatives together: \(f'(x) = h'(g(x)) \, g'(x)\).

This gives you the derivative of the composite function: \(f'(x) = \frac{1}{2}(x-2)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-2}}\). The chain rule helps in efficiently calculating the derivatives even when functions become complex.
Recognizing nested functions and applying this rule is essential for successful differentiation.

Function Domain

Understanding the domain of a function is crucial as it tells us the set of input values (\(x\)) for which the function is defined. In simpler terms, it's all the possible values we can plug into the function without causing mathematical problems, like division by zero or taking the square root of a negative number.
For the function \(f(x) = \sqrt{x-2}\), we need to make sure that the expression inside the square root, \(x-2\), is greater than or equal to zero. This is because the square root of a negative number is not a real number.
Thus, \(x-2\geq0\), solving this inequality gives us \(x \geq 2\). So, the domain of \(f(x)\) is all real numbers where \(x \geq 2\). In interval notation, this is written as \([2, \infty)\).
Knowing the domain helps us understand where the function is valid and can be used to make accurate calculations without errors.

Rationalizing the Denominator

Rationalizing the denominator is a process used in algebra to eliminate radicals (specifically square roots) from the denominator of a fraction. This is often done to make expressions simpler to work with or to conform with standard practice.
Consider the expression \(\frac{1}{\sqrt{2}}\). Having a square root in the denominator is not ideal, so we 'rationalize' it.
To do this, multiply both the numerator and the denominator by the radical in the denominator itself. In this case, that’s \(\sqrt{2}\). So you get:

• Multiply the fraction by \(\frac{\sqrt{2}}{\sqrt{2}}\), which equals 1, thus not changing the value, but changing the appearance.
• The expression becomes \(\frac{1\cdot\sqrt{2}}{\sqrt{2}\cdot\sqrt{2}} = \frac{\sqrt{2}}{2}\).

This technique is frequently applied in calculus and algebraic manipulations to simplify expressions and ensure clarity. When performing calculations, having a rationalized denominator creates cleaner and often more understandable results.