Problem 22 - Precalculus with Limits

Question

In Exercises 17-22, find a formula for the slope of the graph of \(f\) at the point \((x, f(x))\). Then use it to find the slope at the two given points. \(f(x) = \sqrt{x-4}\) (a) \((5, 1)\) (b) \((8, 2)\)

Step-by-Step Solution

Answer

The formula for the slope at any given point \((x, f(x))\) is \(f'(x) = \frac{1}{2\sqrt{x-4}}\). At the point \((5, 1)\), the slope is \(f'(5) = \frac{1}{2}\) and at the point \((8, 2)\), the slope is \(f'(8) = \frac{1}{4}\).

Step 1: Find the derivative (slope) formula

Firstly, rewrite the function \(f(x) = \sqrt{x-4}\) as \(f(x) = (x-4)^{1/2}\). Now, apply the chain rule for differentiation to find \(f'(x)\). The chain rule states that the derivative of a composition of functions is the derivative of the outside function times the derivative of the inside function. The derivative of \(u^{1/2}\) is \(\frac{1}{2}u^{-1/2}\) so we will multiply by the derivative of the inside function (the derivative of \(x-4\) is 1). This gives \(f'(x) = \frac{1}{2}(x-4)^{-1/2} = \frac{1}{2\sqrt{x-4}}\).

Step 2: Find the slope at the point (5, 1)

Substitute the x-coordinate of the first point into the derivative (slope) formula. So at \(x = 5\), the slope is \(f'(5) = \frac{1}{2\sqrt{5-4}} = \frac{1}{2}\).

Step 3: Find the slope at the point (8, 2)

Similarly, substitute the x-coordinate of the second point into the derivative (slope) formula. So at \(x = 8\), the slope is \(f'(8) = \frac{1}{2\sqrt{8-4}} = \frac{1}{4}\).

Key Concepts

Understanding the Chain Rule

In calculus, the chain rule is a fundamental tool used for finding the derivative of a composite function—one function nested within another. For this exercise, for example, we see the square root function encompassing another function \(x-4\). To differentiate \(f(x) = \sqrt{x-4}\) correctly, the function is rewritten as \(f(x) = (x-4)^{1/2}\). This makes it easier to apply the chain rule, which can be summarized in a few simple steps:

• Take the derivative of the outer function: If you have \(g(x) = u^{1/2}\), then the derivative is \(\frac{1}{2}u^{-1/2}\).
• Multiply by the derivative of the inner function: Here, the inside function is \(x-4\) which differentiates to 1.

We then multiply these derivatives together to arrive at the derivative of the composite function. So, applying the chain rule to our function results in \(f'(x) = \frac{1}{2}(x-4)^{-1/2}\), simplifying to \(\frac{1}{2\sqrt{x-4}}\).
The chain rule is particularly useful when tackling complex functions like \(f(x) = (x-4)^{1/2}\) since it simplifies the differentiation process and helps you find the slope of a function more easily.

Slope of a Function

The slope of a function at a given point is represented by the derivative of the function at that point. It tells us how steep the graph of the function is and in which direction it is inclined. In simple terms, the slope is the rate at which y changes with respect to x.
To find the slope of the function at a specific point, we compute the derivative and substitute the x-value of the point into this derivative.

• For point \( (5, 1) \), the slope is found by substituting \(x = 5\) into \(f'(x) = \frac{1}{2\sqrt{x-4}}\) yielding \( \frac{1}{2} \).
• For point \( (8, 2) \), the process is the same, and after substituting \(x = 8\), the slope is \ \frac{1}{4} \.

These results tell us how sharply the curve increases or decreases at these specific points. Understanding this concept is crucial for analyzing the behavior of functions in calculus.

Basics of Calculus Differentiation

Calculus differentiation involves finding the derivative of a function, which essentially represents the rate of change or the slope of that function. This process forms the backbone of what calculus attempts to uncover about various mathematical phenomena. Differentiation lets us determine how functions change and behave over intervals.
In our example, the derivative \(f'(x) = \frac{1}{2\sqrt{x-4}}\) hints at how the original function \(f(x) = \sqrt{x-4}\) changes for different values of \(x\). Techniques like product rule, quotient rule, and chain rule are regularly employed in these processes depending on the complexity and composition of the function.To successfully differentiate:

• Identify what rules apply: For simple functions, use basic rules like power and constant rules.
• Apply these rules to obtain the derivative: Carefully applying the appropriate rule ensures the correct form of the derivative.
• Simplify the derivative function: This makes further analysis or evaluation easier, such as in our exercise where simplifying resulted in \(\frac{1}{2\sqrt{x-4}}\).

Differentiation is a powerful technique used across various scientific and engineering fields to model and predict behaviors and trends.