Problem 7 - Brief Applied Calculus

Question

For each function, find: a. \(f^{\prime \prime}(x)\) and b. \(f^{\prime \prime}(3)\). $$ f(x)=\frac{x-1}{x} $$

Step-by-Step Solution

Answer

a. \( f''(x) = -\frac{2}{x^3} \); b. \( f''(3) = -\frac{2}{27} \).

Step 1: Find the First Derivative

Given the function \( f(x) = \frac{x-1}{x} \), rewrite it as \( f(x) = 1 - \frac{1}{x} \). Now differentiate using the power rule. The derivative \( f'(x) \) of \( 1 \) is \( 0 \), and for \( -\frac{1}{x} \), rewrite \( \frac{1}{x} \) as \( x^{-1} \) and differentiate to get \( f'(x) = 0 - (-1)x^{-2} = \frac{1}{x^2} \). Thus, \( f'(x) = \frac{1}{x^2} \).

Step 2: Find the Second Derivative

Differentiate the first derivative \( f'(x) = \frac{1}{x^2} \). Rewrite this as \( x^{-2} \) and apply the power rule: the derivative of \( x^{-2} \) is \( -2x^{-3} \), which is \( -\frac{2}{x^3} \). Thus, the second derivative \( f''(x) = -\frac{2}{x^3} \).

Step 3: Evaluate the Second Derivative at x = 3

Substitute \( x = 3 \) into the second derivative \( f''(x) = -\frac{2}{x^3} \). This gives \( f''(3) = -\frac{2}{3^3} = -\frac{2}{27} \).

Key Concepts

Derivatives

In calculus, derivatives are critical concepts used to understand how a function changes as its input changes. Imagine you're driving a car, and you want to know your speed at a specific moment. This is similar to finding the derivative of your position function with respect to time.

• The **first derivative** of a function gives the *rate of change* or *slope* of the function at any given point.
• For a function \( f(x) \), the first derivative, denoted as \( f'(x) \) or \( \frac{d}{dx}f(x) \), tells you how \( f(x) \) changes with respect to \( x \).
• In the original exercise, the function \( f(x) = \frac{x-1}{x} \) was first rewritten as \( f(x) = 1 - \frac{1}{x} \) before finding its derivative. This is because it's often easier to work with polynomial expressions when differentiating.

Understanding derivatives allows us to predict and analyze the behavior of functions, aiding in problems that involve motion, growth, and change.

Power Rule

The power rule is one of the simplest and most often used rules in differentiation. It provides a straightforward method to find the derivative of power functions, which are essentially functions where a variable is raised to a constant power.

• The power rule states that if you have a function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \).
• In our exercise, to find the derivative of \( \frac{1}{x} \), it was rewritten as \( x^{-1} \). Applying the power rule gives \( -1x^{-2} \), or \( -\frac{1}{x^2} \).
• Similarly, when the first derivative \( \frac{1}{x^2} \) was differentiated again using the power rule, \( f''(x) \) was found to be \( -\frac{2}{x^3} \).

This rule is powerful because it simplifies the process of differentiation, cutting through complex algebraic manipulation with a simple formula: multiply by the power, then decrease the power by one.

Second Derivative

The second derivative of a function provides even more insight into the behavior of the function, particularly concerning its *concavity* and *inflection points*.

• If the first derivative \( f'(x) \) measures the rate of change, the **second derivative** \( f''(x) \) measures the rate at which that rate of change is itself changing.
• A positive second derivative indicates that the function is concave upwards, while a negative second derivative indicates concave downwards.
• In our exercise, after calculating \( f''(x) = -\frac{2}{x^3} \), we evaluated it at \( x = 3 \) to find \( f''(3) = -\frac{2}{27} \). This negative value tells us that the function is concave down at \( x = 3 \).

Understanding second derivatives is crucial in many applications, such as finding acceleration in physics or determining the curvature of graphs in optimization problems. They reveal deeper characteristics about the behavior of functions beyond what the first derivative can show.