Problem 26 - Calculus - AP Edition

Question

Approximate \(f^{\prime}(1)\) by considering the difference quotient \(\quad \frac{f(1+h)-f(1)}{h}\) for values of \(h\) near \(0,\) and then find the exact value of \(f^{\prime}(1)\) by differentiating. $$f(x)=\frac{1}{x^{2}}$$

Step-by-Step Solution

Answer

The exact value of \( f'(1) \) is \(-2\).

Step 1: Apply the Difference Quotient

We need to approximate \(f'(1)\) by using the difference quotient \( \frac{f(1+h)-f(1)}{h} \). The function \( f(x) = \frac{1}{x^2} \) implies \( f(1) = 1 \). Substituting \( f(x) \) into the difference quotient, we have \( \frac{\frac{1}{(1+h)^2} - 1}{h} \).

Step 2: Simplify the Expression

Simplify the expression \( \frac{\frac{1}{(1+h)^2} - 1}{h} \). The expression becomes \( \frac{1 - (1+h)^2}{h(1+h)^2} \) after combining to a common denominator and expanding \((1+h)^{-2}\).

Step 3: Evaluate at Small h

Evaluate \( \frac{1 - (1+h)^2}{h(1+h)^2} \) by letting \( h \) approach small values such as \( 0.1, 0.01, \) and \( 0.001 \) to approximate \(f'(1)\). Each evaluation gets us closer to the slope, suggesting that \(f'(1) \approx -2\).

Step 4: Differentiate the Function

Determine the exact derivative of the function \( f(x) = \frac{1}{x^2} \). Using the power rule on \( f(x) = x^{-2} \), we find \( f'(x) = -2x^{-3} \).

Step 5: Evaluate the Derivative at x = 1

Substitute \( x = 1 \) into the derivative \( f'(x) = -2x^{-3} \). This gives \( f'(1) = -2(1)^{-3} = -2 \). Thus, the exact value of \( f'(1) \) is \(-2\).

Key Concepts

Difference Quotient

When we're talking about the difference quotient, we're dealing with the fundamental building block of calculus: the derivative. The difference quotient is a formula used to approximate the slope of the tangent line at a particular point on a curve. For a given function \( f(x) \), the difference quotient is expressed as \( \frac{f(x+h)-f(x)}{h} \), where \( h \) is a small increment in \( x \).

This formula allows us to calculate the average rate of change of the function over an interval. For small values of \( h \), this average change begins to approximate the instantaneous rate of change at \( x \). Thus, calculating the difference quotient for multiple small values of \( h \) closely approximates the derivative at a point.

• The numerator \( f(x+h) - f(x) \) gives the change in the function's value.
• The denominator \( h \) is the change in \( x \) itself.

This simple yet powerful process is a stepping stone into the exact derivative found using calculus techniques like analytical differentiation.

Approximation

Approximation in calculus is identifying the behavior or value of a function at a specific point using estimated methods when exact analytical expressions are tricky or infeasible. One common way to approximate derivatives is through the difference quotient with smaller and smaller \( h \).

During the approximation process, we perform calculations for several values of \( h \) approaching zero, such as \( 0.1, 0.01, \) and \( 0.001 \), and observe how the results converge to reveal the function's actual derivative. These calculations provide a picture of the function's behavior around a point, highlighting:

• The slope of the tangent line being approached.
• The precision of approximation improves with smaller \( h \).

It's essential to understand that approximations are an essential part of doing calculus practically, especially when dealing with complex functions or real-world data that are hard to differentiate outright.

Derivative

The derivative of a function measures how the function's output changes as its input changes. It's the core concept that captures the idea of a rate of change or slope of the function at a specific point.

To find the exact derivative of a function like \( f(x) = \frac{1}{x^2} \), we go beyond approximation and use rules of differentiation. Here, the power rule comes handy. We rewrite \( f(x) \) as \( x^{-2} \) and differentiate using the power rule, yielding \( f'(x) = -2x^{-3} \). This reveals exactly how steep the function is at any point \( x \).

• The derivative at a particular point, such as \( x = 1 \), gives us a precise rate of change, which means if \( f'(1) = -2 \), the function decreases by 2 units for every unit increase in \( x \) at that point.
• Understanding derivatives is crucial for solving problems involving changes and predicting future behavior in fields like physics, engineering, and economics.

Thus, learning how to compute and interpret derivatives enables you to unlock many real-world and mathematical insights.