Problem 5 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 5 through 8 , estimate \(f^{\prime}(c)\) by calculating the difference quotient \(\frac{f(c+h)-f(c)}{h}\) for successively smaller values of \(|h| .\) Use both positive and negative values of \(h .\). $$ f(x)=\sqrt{x} \text { . Approximate } f^{\prime}(9) \text { . } $$

Step-by-Step Solution

Answer

The approximate value of the derivative of the function \(f(x) = \sqrt{x}\) at \(x = 9\), denoted as \(f'(9)\), can be estimated by considering positive and negative differences of \(h\) approaching 0. You will compute \(\frac{\sqrt{9 + h} - 3}{h}\) and \(\frac{\sqrt{9 - h} - 3}{h}\) while letting \(h\) approach 0 from both positive and negative directions respectively.

Step 1: Compute f(c+h) and f(c)

First, calculate the values of \(f(c+h)\) and \(f(c)\). Here, \(c = 9\), and \(h\) is a small number, which can be positive or negative. To calculate \(f(c+h)\), replace \(x\) in \(f(x) = \sqrt{x}\) by \(9 + h\), and calculate \(f(9 + h) = \sqrt{9 + h}\). Similarly, \(f(9) = \sqrt{9} = 3\).

Step 2: Compute the Difference Quotient

The difference quotient is defined as \(\frac{f(c+h)-f(c)}{h}\). So, substitute the previously calculated values into the formula. For positive \(h\), the difference quotient would be \(\frac{\sqrt{9 + h} - 3}{h}\), and for negative \(h\) it would be \(\frac{\sqrt{9 - h} - 3}{h}\).

Step 3: Approximate f'(c)

Estimate \(f'(9)\) by choosing smaller and smaller absolute values of \(h\) (both positive and negative), and calculating the difference quotient for each value. The limit as \(h\) approaches 0 of these difference quotients would give the approximation of \(f'(9)\).

Key Concepts

Derivative Approximation

The concept of derivative approximation involves calculating the derivative of a function at a certain point by using numerical methods instead of exact calculations. When dealing with functions like square roots, finding the derivative directly might seem challenging at first. Instead, one useful approach is to use the **difference quotient**. This quotient is given by the formula:

• \( \frac{f(c+h) - f(c)}{h} \)

Here, \( f(c+h) \) is the function value close to \( c \), and \( f(c) \) is the function value at \( c \). By computing this difference quotient for progressively smaller values of \(|h|\), and using values both above and below \( c \), we can get a good approximation for the derivative. As \( h \) becomes minuscule, this quotient approaches the actual derivative, \( f'(c) \). This is crucial for functions where calculating the derivative is not straightforward analytically.

Limits

The concept of limits is fundamental when you're dealing with derivatives and especially when finding them through approximation. In the case of the difference quotient approach, as \( h \) gets closer and closer to zero, the quotient

• \( \frac{f(c+h) - f(c)}{h} \)

begins to converge to the function's true derivative at that point. Limits allow us to make sense of these values trending towards something meaningful, even though \( h \) is not zero. This is because division by zero isn't defined. But with limits, we don't need \( h \) to be zero, just very close! This makes limits a powerful tool in calculus. In exercises where you approximate derivatives, thinking of the difference quotient's behavior as \( h \) shrinks helps you understand how we arrive at the derivative. This gives insight into how the tangent line's slope at \( c \) is "born" from these values farming "infinitely" near \( c \).

Square Roots

In this problem, the function in question includes a square root, \( f(x) = \sqrt{x} \). Functions with square roots can appear daunting, particularly when differentiating. They require special attention, considering their domain limits, since you can't take square roots of negative numbers in real-number analysis. When calculating a derivative or a difference quotient, it's important to ensure that the values set for \( c+h \) remain non-negative.What's particularly useful about square root functions in derivatives is understanding their behavior: they grow steadily slower as \( x \) increases. This slowing slope can often be easily sensed if you imagine the curve flattening out. By focusing on approximating derivatives at specific points, like \( c = 9 \), we can get a sense of this slowdown via numerical differences from slightly adjusted points surrounding \( c \). Understanding these subtleties is part of why calculus can feel like both an art and a science, revealing the subtleties of how functions change and develop.