Problem 76 - Calculus Single Variable

Question

Approximate \(f^{\prime}(c)\) for the given \(f\) and \(c\) in the following way: find a small viewing window with \(P=\) \((c, f(c))\) near the center. The window should be small enough so that the graph of \(f\) appears to be a straight line. Let \(Q\) and \(R\) be the endpoints of the graph of \(f\) as it exits this window. Use the slope of \(Q R\) as the approximation to \(f^{\prime}(c)\). $$ f(x)=x-1 / x, \quad c=1 $$

Step-by-Step Solution

Answer

The approximation of \( f'(1) \) is 2.0425.

Step 1: Determine the Value of f(c)

Calculate the value of the function \( f(x) = x - \frac{1}{x} \) at \( c = 1 \). Substituting \( c = 1 \) into the function, we get \( f(1) = 1 - \frac{1}{1} = 0 \). Thus, the point \( P \) is \( (1, 0) \).

Step 2: Choose a Viewing Window

Choose a small window around \( c = 1 \) to ensure the graph of \( f \) looks linear. A convenient small window can be \([0.8, 1.2]\). This means we will examine the behavior of the function from \( x = 0.8 \) to \( x = 1.2 \).

Step 3: Find Points Q and R

The point \( Q \) will correspond to \( x = 0.8 \) and \( R \) to \( x = 1.2 \). Calculate \( f(0.8) = 0.8 - \frac{1}{0.8} = 0.8 - 1.25 = -0.45 \) and \( f(1.2) = 1.2 - \frac{1}{1.2} = 1.2 - 0.833 \approx 0.367 \). Thus, \( Q = (0.8, -0.45) \) and \( R = (1.2, 0.367) \).

Step 4: Calculate Slope of QR

The slope of the line segment \( QR \) is given by: \[ m = \frac{f(1.2) - f(0.8)}{1.2 - 0.8} = \frac{0.367 - (-0.45)}{1.2 - 0.8} = \frac{0.367 + 0.45}{0.4} = \frac{0.817}{0.4} = 2.0425 \]. Therefore, the approximate value of \( f'(1) \) is 2.0425.

Key Concepts

Slope of a Secant Line

When studying calculus, understanding what the slope of a secant line is can be incredibly beneficial. Imagine you're drawing a straight line over a curve: that straight line is called a secant line. In our exercise, the secant line connects two points, \( Q \) and \( R \), on the curve of the function \( f(x) \). The secant line's slope approximates the derivative at a point.
This line essentially gives us a simplified idea of how steep the curve is between two points. To find this slope:

• Determine the difference in \( y \)-values (the outputs of the function).
• Divide by the difference in \( x \)-values (the inputs).

In our specific problem, this computation is done by finding the outputs at \( x = 0.8 \) and \( x = 1.2 \), which correspond to points \( Q \) and \( R \). Then, you calculate the slope to find an approximation to the derivative \( f^{\prime}(c) \). This gives us insight into how the function acts around \( c = 1 \).

Function Evaluation

Evaluating a function is a foundational skill in calculus and algebra. It involves calculating the output of a function for a specific input, or \( x \)-value. In this exercise, we begin by evaluating the function \( f(x) = x - \frac{1}{x} \) at \( c = 1 \). This gives us our initial point of interest, \( P = (1, 0) \).
Function evaluation is essentially about substituting the given \( x \) value into the function's formula and simplifying to find the result. For instance:

• For \( x = 0.8 \), calculate \( f(0.8) \) to find the \( y \)-value at that point.
• Similarly, for \( x = 1.2 \), compute \( f(1.2) \) to find another \( y \)-value.

Evaluating the function at these points helps in understanding how the function behaves around the point \( c = 1 \), particularly when establishing the slope of the secant line.

Viewing Window

In order to approximate derivatives effectively, selecting the correct viewing window is crucial. The viewing window is the range of \( x \)-values we focus on to simplify and effectively approximate the function. In this exercise, a small window of \([0.8, 1.2]\) was selected.
The purpose of choosing a small viewing window is to make the graph of the function appear linearly over that range. This means:\

• The function graph looks straight.
• We can use a straight line (the secant) to approximate changes in the function.

This approach aids in seeing the behavior of \( f(x) \) clearly without the complexities that larger ranges might introduce. With a carefully chosen range, you can gain a clearer insight into how the function approximates linear behavior around the specific point of interest.

Linear Approximation

Linear approximation involves modeling a function's behavior with a straight line over a small interval. This is a powerful concept used to navigate complex functions by making them simpler. By understanding that small sections of differentiable functions can behave linearly, we can approximate and understand them better.
In the context of our problem:

• The secant line \( QR \) acts as a simple linear approximation of the function at \( c = 1 \).
• Despite the overall curve, the section between \( Q \) and \( R \) forms a relatively straight line, making computations straightforward.

By using the linear approximation, you get a glimpse into how the function behaves at and near a specific point. If you think about it, linear approximation simplifies our world, letting us use the straightforward properties of lines to manage the complexity of curves. This approximation is critical when visual aids like graphs make understanding derivatives more intuitive.