Problem 21 - 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. \(f(x) = \sqrt{x-1}\) (a) \((5, 2)\) (b) \((10, 3)\)

Step-by-Step Solution

Answer

The slopes of the function \(f(x) = \sqrt{x-1}\) at the points (5, 2) and (10, 3) are \(\dfrac{1}{4}\) and \(\dfrac{1}{6}\), respectively.

Step 1: Differentiate the function

Apply the chain rule for differentiation. Let \(u(x) = x - 1\) and \(v(u) = \sqrt{u}\). The derivative of \(v(u)\) is \( \dfrac{1}{2 \sqrt{u}}\). So, by the chain rule, \(f'(x) = v'(u(x)) \cdot u'(x) = \dfrac{1}{2 \sqrt{x-1}} \cdot 1\). This simplifies to \(f'(x) = \dfrac{1}{2 \sqrt{x-1}}\).

Step 2: Find the slope at the first point

Substitute \(x = 5\) into the derivative to obtain the slope at the first point. Thus, \(f'(5) = \dfrac{1}{2 \sqrt{5-1}} = \dfrac{1}{2 \cdot 2} = \dfrac{1}{4}\).

Step 3: Find the slope at the second point

Now substitute \(x = 10\) into the derivative to find the slope at the second point. This yields \(f'(10) = \dfrac{1}{2 \sqrt{10-1}} = \dfrac{1}{6}\).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus used to find derivatives of composite functions. To understand it, think of it as breaking down the process of differentiation into parts. When a function is composed of two or more functions, we can use the chain rule to differentiate it efficiently.

In our exercise, the function is given as \( f(x) = \sqrt{x-1} \). Here, instead of directly differentiating, we recognize this as a composition of two functions:

• The inside function is \( u(x) = x - 1 \).
• The outside function takes this result, so \( v(u) = \sqrt{u} \).

According to the chain rule, the derivative of \( f(x) \) is found by multiplying the derivative of \( v(u) \) by the derivative of \( u(x) \). So, the formula is:

\[f'(x) = v'(u(x)) \cdot u'(x)\]

In this case:

• Derive \( v(u) \): \( v'(u) = \frac{1}{2\sqrt{u}} \)
• Derive \( u(x) \): \( u'(x) = 1 \)

Putting it all together, we have \( v'(u) \cdot u'(x) = \frac{1}{2 \sqrt{x-1}} \cdot 1 \). This gives us the derivative \( f'(x) = \frac{1}{2\sqrt{x-1}} \).

Slope of a Curve

Finding the slope of a curve at a point involves determining how steep the curve is at that specific location. The slope of the graph of a function at any given point is essentially the derivative evaluated at that point.

For our exercise, we need to find the derivative of \( f(x) = \sqrt{x-1} \), which we already computed as \( f'(x) = \frac{1}{2\sqrt{x-1}} \). To find the slope at a specific point, such as \((5, 2)\) or \((10, 3)\), we simply plug the \(x\)-value from this point into the derivative:

• Point (5, 2): Substitute \(x = 5\) into the derivative. The slope \( f'(5) = \frac{1}{2 \sqrt{5-1}} = \frac{1}{4} \).
• Point (10, 3): Substitute \(x = 10\) into the derivative. The slope \( f'(10) = \frac{1}{2 \sqrt{10-1}} = \frac{1}{6} \).

Thus, at each point, the calculated derivative represents the slope of the tangent line to the curve at that point.

Differentiation

Differentiation is the process of finding the derivative of a function. It is a way to calculate the rate at which a function is changing at any given point. This concept is central to calculus and helps determine properties like the slope, concavity, and other behavior of functions.

In the context of the given function \( f(x) = \sqrt{x-1} \), differentiation involves applying the chain rule, as explained. By taking the derivative, we find out how the function behaves around any point and how it responds to changes in \(x\).

The significance of differentiation lies in:

• Understanding the rate of change: This means knowing how fast or slow something is changing.
• Finding tangents: Differentiation gives us the slope of the tangent line at any point.
• Modeling real-life problems: Many physical processes can be modeled using differentiable functions.

Therefore, knowing how to differentiate and apply it helps in understanding mathematical models and real-world problems more deeply.