Problem 3 - Active Calculus

Question

Suppose that \(f(x)\) is a function with \(f(125)=76\) and \(f^{\prime}(125)=8\). Estimate \(f(128.5)\). \(f(128.5)=\) \(\square\)

Step-by-Step Solution

Answer

f(128.5) \approx 104.

Step 1: Identify the given information

You are given that at the point where \( f(125) = 76 \) and \( f^{\text{‘}}(125) = 8 \). You need to estimate \( f(128.5) \).

Step 2: Recognize the use of linear approximation

To estimate \( f(128.5) \), use the linear approximation formula: \[ f(x) \approx f(a) + f^{‘}(a) (x - a) \]. Here, \( a = 125 \) and \( x = 128.5 \).

Step 3: Substitute the given values

Substitute \( a = 125 \), \( x = 128.5 \), \( f(125) = 76 \), and \( f^{‘}(125) = 8 \) into the linear approximation formula: \[ f(128.5) \approx 76 + 8 \times (128.5 - 125) \].

Step 4: Simplify the expression

Calculate the difference and multiply: \[ 128.5 - 125 = 3.5 \] \[ 8 \times 3.5 = 28 \]. So, \[ f(128.5) \approx 76 + 28 \].

Step 5: Add the results to find the estimate

Finally, add the values: \[ 76 + 28 = 104 \]. Therefore, \[ f(128.5) \approx 104 \].

Key Concepts

Differential Calculus

Differential calculus focuses on the idea of the derivative, which represents the rate of change of a function. Essentially, it's about understanding how a function changes as its input changes. The derivative of a function, denoted as \( f^{\text{‘}}(x) \), indicates how fast or slow the function \( f(x) \) is increasing or decreasing at any point \( x \).

In the given problem, \( f^{\text{‘}}(125) = 8 \) implies that at \( x = 125 \), the function \( f(x) \) increases by 8 units for every 1 unit increase in \( x \). This information is essential for estimating function values close to \( x = 125 \). Differential calculus helps us calculate these changes accurately using the concept of linear approximation.

Tangent Line Approximation

Tangent line approximation, also known as linear approximation, helps us estimate the value of a function close to a known point. Imagine drawing a straight line that just touches the curve of the function at a given point – this line is the tangent line. The slope of this line is the derivative \( f^{\text{‘}}(x) \) of the function at that point.

The main idea is that if you move a little from the known point along the x-axis, the value of the function at the new point can be approximated using the tangent line. The formula we use for tangent line approximation is:
\[ f(x) \approx f(a) + f^{\text{‘}}(a)(x - a) \] where \( a \) is the known point, and \( f(a) \) and \( f^{\text{‘}}(a) \) are the function value and its derivative at that point, respectively.

Using this method, we accurately estimate function values for points close to \( a \). For example, in our problem, we used this formula to approximate \( f(128.5) \) using the known values at \( x = 125 \). This gives us a quick and efficient way to estimate without calculating the exact function value.

Estimating Function Values

Estimating function values using differential calculus and tangent line approximation is both powerful and practical. In many real-world scenarios, it's not feasible to compute the exact values of functions due to complexity or limited information.

Here, we used the data points \( f(125) = 76 \) and \( f^{\text{‘}}(125) = 8 \) to estimate \( f(128.5) \). By applying the linear approximation formula:
\[ f(128.5) \approx 76 + 8 \times (128.5 - 125) \] we simplify it to:
\[ 76 + 8 \times 3.5 = 104 \]
This technique of linear approximation is extremely useful because:

• It's quick and easy to use for practical purposes.
• It allows us to gauge the function's behavior near a known point.
• It provides reasonable estimates with less computational effort.

Remember, this method works best for points close to \( a \), and the further away you get, the less accurate the approximation might be. However, it's a great way to make informed guesses about function values in everyday situations!