Problem 9 - Precalculus: Mathematics for Calculus

Question

Find the slope of the tangent line to the graph of \(f\) at the given point. $$f(x)=\frac{5}{x+2}, \quad \text { at }(3,1)$$

Step-by-Step Solution

Answer

The slope of the tangent line at (3,1) is \(-\frac{1}{5}\).

Step 1: Understand the Problem

We need to find the slope of the tangent line to the function \( f(x) = \frac{5}{x+2} \) at the point \((3,1)\). The slope of the tangent line is given by the derivative of the function evaluated at that point.

Step 2: Find the Derivative of the Function

To find the slope of the tangent line, we differentiate the function \( f(x) = \frac{5}{x+2} \). Using the quotient rule or rewriting \( f(x) = 5(x+2)^{-1} \) and applying the power rule, we find the derivative: \[f'(x) = -5(x+2)^{-2} = \frac{-5}{(x+2)^2}.\]

Step 3: Evaluate the Derivative at the Given Point

Now substitute \( x = 3 \) into the derivative \( f'(x) = \frac{-5}{(x+2)^2} \) to find the slope of the tangent line at \( (3,1) \):\[f'(3) = \frac{-5}{(3+2)^2} = \frac{-5}{25} = -\frac{1}{5}.\]

Step 4: Interpret the Result

The slope of the tangent line to the graph of the function at the point \((3,1)\) is \(-\frac{1}{5}\). This implies that the tangent line is decreasing at this point.

Key Concepts

Derivative

Understanding derivatives is central to calculus and is essential for finding the slope of a tangent line. The derivative of a function represents how the function's value changes as its input changes. It's like measuring how fast something is moving at an exact moment rather than over a period of time.

For a function like \( f(x) = \frac{5}{x+2} \), the derivative tells us how the function's value will change with a very tiny change in \( x \). Calculating the derivative using rules like the power rule or quotient rule helps us find this rate of change.

• **Power Rule**: Applies when differentiating terms like \( x^n \).
• **Quotient Rule**: Used for dividing functions, which we'll explore later.

These rules help break down complex calculations into simpler steps, making it easier to find derivatives quickly. The derivative of our function, \( f'(x) = \frac{-5}{(x+2)^2} \), captures the rate at which \( f(x) \) changes at any point \( x \).

Slope

In the context of a tangent line, the slope is a measure of how steep the line is at a specific point on the graph. Think of it as the tilt or angle of the line. For straight lines, finding the slope is straightforward, but for curves, it requires calculus and derivatives.

The slope of a tangent line to a curve at any given point is precisely the derivative evaluated at that point. This tells us how quickly the function is increasing or decreasing there.

• A positive slope means the function is increasing.
• A zero slope indicates a temporary pause in change.
• A negative slope, like \(-\frac{1}{5}\) in our problem, shows a decrease.

This slope helps in constructing the tangent line equation and provides insights into the function's behavior at that precise location.

Quotient Rule

The quotient rule is a critical tool when dealing with derivatives in functions that are fractions of two other functions. It helps find the derivative of a quotient of two differentiable functions expressed as \( \frac{u}{v} \).

The rule is structured as follows:

• Let \( u(x) \) and \( v(x) \) be functions.
• The derivative \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

This formula breaks the problem into manageable parts by focusing on the numerator and denominator's rates of change separately. In our example, rewriting \( f(x) \) as \( 5(x+2)^{-1} \) was an alternative, but understanding the quotient rule can also provide another perspective on solving the same problem more systematically for different expressions.