Problem 46 - Calculus Early Transcendentals

Question

Use the definition of a derivative to find \(f^{\prime}(x)\) and \(f^{\prime \prime}(x)\) . Then graph \(f, f^{\prime},\) and \(f^{\prime \prime}\) on a common screen and check to see if your answers are reasonable. \(f(x)=1 / x\)

Step-by-Step Solution

Answer

First derivative: \(-\frac{1}{x^2}\), second derivative: \(\frac{2}{x^3}\). Graphs confirm expected behaviors.

Step 1: Write the Definition of the Derivative

The derivative of a function \( f(x) \) is defined as \[ f^{\prime}(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \] We will use this definition to find \( f^{\prime}(x) \) for \( f(x) = \frac{1}{x} \).

Step 2: Substitute \( f(x) = \frac{1}{x} \) into the Definition

Substitute \( f(x) \) and \( f(x+h) \) into the derivative formula:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \] This simplifies to:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{x - (x+h)}{h(x+h)x} \] \[ f^{\prime}(x) = \lim_{h \to 0} \frac{-h}{h(x+h)x} \]

Step 3: Simplify the Expression and Take the Limit

Cancel \( h \) from the numerator and the denominator:\[ f^{\prime}(x) = \lim_{h \to 0} \frac{-1}{(x+h)x} \] Evaluate the limit as \( h \to 0 \):\[ f^{\prime}(x) = \frac{-1}{x^2} \] Thus, the first derivative is \( f^{\prime}(x) = -\frac{1}{x^2} \).

Step 4: Find the Second Derivative

Now, we'll find the derivative of \( f^{\prime}(x) = -\frac{1}{x^2} \):\[ f^{\prime\prime}(x) = \frac{d}{dx}\left(-\frac{1}{x^2}\right) \] Using the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \), we get:\[ f^{\prime\prime}(x) = 2x^{-3} \]Thus, \( f^{\prime\prime}(x) = \frac{2}{x^3} \).

Step 5: Graph the Functions

Graph the original function \( f(x) = \frac{1}{x} \), the first derivative \( f^{\prime}(x) = -\frac{1}{x^2} \), and the second derivative \( f^{\prime\prime}(x) = \frac{2}{x^3} \) on the same set of axes. Check if the graphs align with the expected behavior: - \( f(x) \) is a hyperbola symmetric about the origin.- \( f^{\prime}(x) \) shows a negative slope, reflecting that \( f(x) \) is decreasing.- \( f^{\prime\prime}(x) \) shows positive curvature, indicating \( f(x) \) is concave up.

Key Concepts

Limit Definition of Derivative

The derivative of a function provides us with the rate at which the function is changing at any given point. To obtain the derivative of a function, we often use the limit definition of derivative. This is a fundamental concept in calculus that allows us to find the slope of the tangent line to a curve at a specific point. The formal definition is given by the formula:

• \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

Here, \( h \) represents a small increment in \( x \), and as \( h \) approaches 0, the difference quotient approaches the derivative \( f'(x) \). This formula is essential to derive the first and second derivatives of a function. In our example, we used this definition to find the derivative of \( f(x) = \frac{1}{x} \). The goal is to express \( f'(x) \) as a limit and evaluate it as \( h \to 0 \). It’s a critical technique for calculating exact rates of change mathematically.

First Derivative

The first derivative of a function \( f(x) \), denoted \( f'(x) \), is the result of applying the derivative process to \( f(x) \). It provides important information about the function's rate of change. For the function \( f(x) = \frac{1}{x} \), the first derivative is calculated as follows:

• Start with \[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{x+h} - \frac{1}{x}}{h} \]
• Simplify the expression: \[ f'(x) = \lim_{h \to 0} \frac{-h}{h(x+h)x} \]
• Cancel \( h \): \[ f'(x) = \lim_{h \to 0} \frac{-1}{(x+h)x} \]
• Evaluate the limit: \[ f'(x) = -\frac{1}{x^2} \]

The first derivative \( f'(x) = -\frac{1}{x^2} \) tells us that the function \( f(x) \) is decreasing, as indicated by the negative sign. The magnitude of \( f'(x) \) also gives us insight into the steepness of the curve at various points.

Second Derivative

The second derivative is the derivative of the first derivative, providing information about the curvature of the original function. It's denoted as \( f''(x) \) and shows how the rate of change of the function's slope itself changes. For \( f(x) = \frac{1}{x} \), the second derivative is found by differentiating \( f'(x) = -\frac{1}{x^2} \):

• Using the power rule for differentiation: \[ f''(x) = \frac{d}{dx} \left(-\frac{1}{x^2}\right) \]
• Simplify with the rule \( \frac{d}{dx}[x^n] = nx^{n-1} \): \[ f''(x) = 2x^{-3} \]
• Express properly: \[ f''(x) = \frac{2}{x^3} \]

The second derivative \( f''(x) = \frac{2}{x^3} \) indicates concavity. The positive value suggests that the function \( f(x) \) is concave upwards. This can help in determining the nature of stationary points—whether they are maxima, minima, or points of inflection.

Graphing Derivatives

Graphing a function along with its derivatives helps visually understand the relationship between a function and its rate of change. Here are some key ideas for graphing \( f(x) = \frac{1}{x} \), \( f'(x) = -\frac{1}{x^2} \), and \( f''(x) = \frac{2}{x^3} \):

• The original function \( f(x) \) is a hyperbola centered around the origin, representing a rapidly decreasing trend in both quadrants.
• The first derivative \( f'(x) \) graph shows a smooth, consistently negative curve, affirming that \( f(x) \) is always decreasing.
• Finally, the second derivative \( f''(x) \) reflects a graph that remains above the \( x \)-axis, indicating positive curvature—which means \( f(x) \) is concave up.

By graphing these functions together, we can verify the accuracy of our derivatives and visually comprehend how the function's slope and curvature change across the domain. This makes the conceptual understanding of calculus comprehensive and grounded.