Problem 32 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(f\) is a function satisfying \(f(1)=10\), and \(f^{\prime}(1.02)=12 .\) Use this information to approximate \(f(1.02) .\)

Step-by-Step Solution

Answer

The approximate value of \(f(1.02)\) is 10.24.

Step 1: Understanding the Problem

We are given a function \(f\) with a known value at \(x=1\), which is \(f(1) = 10\), and its derivative at \(x = 1.02\) is \(f'(1.02) = 12\). We need to approximate the value of \(f(1.02)\).

Step 2: Using Linear Approximation Formula

We can use the linear approximation formula, which states that for a function \(f\) and small \(\Delta x\), \(f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x\). Here, \(x = 1\), \(\Delta x = 1.02 - 1 = 0.02\).

Step 3: Applying the Values to the Formula

Substitute \(x = 1\), \(f(1) = 10\), \(f'(1.02) = 12\), and \(\Delta x = 0.02\) into the linear approximation formula:\[ f(1.02) \approx f(1) + f'(1.02) \cdot (1.02 - 1) \]Substitute the known values:\[ f(1.02) \approx 10 + 12 \cdot 0.02 \]

Step 4: Performing the Calculation

Calculate the expression:\[ f(1.02) \approx 10 + 12 \cdot 0.02 \]\[ f(1.02) \approx 10 + 0.24 \]\[ f(1.02) \approx 10.24 \]

Step 5: Concluding the Solution

The approximate value for \(f(1.02)\) is 10.24, using the linear approximation technique based on the provided function values and its derivative.

Key Concepts

Differential Calculus

Differential calculus is a branch of mathematics that deals with the concept of how things change. It provides us with tools to find rates of change, which is crucial for reflecting how a function behaves at a specific point.
In simpler terms, differential calculus focuses on understanding how a small change in one variable affects another. This is often depicted through derivatives, which help in finding these rates of change for any given function.

• The main focus is on finding the derivative, which represents the slope of the function at any given point.
• Derivatives help us analyze how a change in input values results in a change in the output.

For example, if we know the derivative of a function, we can predict how the function will change near a specific point. This is particularly useful in physics and engineering, where understanding precise rates of change can have significant implications.

Derivative at a Point

The derivative of a function at a specific point gives us the exact rate of change or slope of the function at that point. This is like zooming in on a curve to see how steep it is right there.
Using the derivative at a particular point, say at \(x = a\), we can express this as \(f'(a)\). What this means is that if you slightly change \(x\) around \(a\), the function \(f\) will change by about \(f'(a) \times \Delta x\).

• A positive derivative indicates the function is increasing at that point.
• A negative derivative implies the function is decreasing.
• A zero derivative often means the function might have a local maximum or minimum.

In the given problem, we used \(f'(1.02) = 12\), which tells us not just the slope at \(x = 1.02\) but also helps us to make approximations and predictions about the function's behavior near this point.

Function Approximation

Function approximation, particularly through linear approximation, allows us to estimate the value of a function near a certain point based on known information. This technique is particularly useful when a function might be complex, and we need to understand its behavior locally without having a detailed formula.
The linear approximation formula is given by:\[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x \]It uses the derivative to "approximate" how the function behaves close to \(x\).

• \(f(x)\) is the value of the function at a known point.
• \(f'(x)\) is the rate of change of the function at \(x\).
• \(\Delta x\) is a small change in \(x\), meaning how far we are moving from the known point.

This approach is extremely useful when you have initial data and need to predict or estimate values, without recalculating the entire function from scratch. The approximate value of \(f(1.02) = 10.24\) was determined using this very approach.