Problem 75 - 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 \(f^{\prime}(c)\). $$ f(x)=\left(x^{2}+x+1\right) /\left(x^{2}+2\right), c=1 $$

Step-by-Step Solution

Answer

The approximate derivative \( f'(1) \) is 0.335.

Step 1: Define Points P, Q, and R

First, calculate the point \( P \) which is \((c, f(c))\). Here, \( c = 1 \), so we compute \( f(1) = \frac{1^2 + 1 + 1}{1^2 + 2} = \frac{3}{3} = 1 \). Thus, \( P = (1, 1) \). Then, choose small increments \( \Delta x \) around \( c = 1 \), such as \( x = 0.9 \) and \( x = 1.1 \), to find points \( Q \) and \( R \).

Step 2: Calculate Function Values at Q and R

Calculate \( f(0.9) \) and \( f(1.1) \). For \( Q \): \( f(0.9) = \frac{0.9^2+0.9+1}{0.9^2+2} = \frac{2.71}{2.81} \approx 0.964 \). For \( R \): \( f(1.1) = \frac{1.1^2+1.1+1}{1.1^2+2} = \frac{3.31}{3.21} \approx 1.031 \). Thus, \( Q = (0.9, 0.964) \) and \( R = (1.1, 1.031) \).

Step 3: Calculate the Slope of QR

Use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) to find the slope of the line \( QR \). Here, \( (x_1, y_1) = (0.9, 0.964) \) and \( (x_2, y_2) = (1.1, 1.031) \). Thus, \( m = \frac{1.031 - 0.964}{1.1 - 0.9} = \frac{0.067}{0.2} = 0.335 \).

Step 4: Approximate Derivative at c

The slope of line \( QR \) approximates \( f'(c) \). Therefore, the derivative \( f'(1) \approx 0.335 \).

Key Concepts

Understanding Derivatives in Calculus

Derivatives are a fundamental concept in calculus. They represent the rate at which a function's value changes as its input changes. You can think of a derivative as a way to predict how a curve behaves just by looking at a particular point on it.
The notation for a derivative, such as \( f'(x) \), indicates the slope of the tangent line to the function \( f \) at the point \( x \). This "slope" speaks to how much the function \( f \) is increasing or decreasing at that point.
In essence, when you compute a derivative, you're finding a formula that gives you this slope for any value of \( x \). This understanding is crucial for predicting trends, optimizing outcomes in various fields, and analyzing dynamic systems.

Concept of Tangent Slope

In geometry, a tangent line is a straight line that just "touches" a curve at one point. The slope of this tangent line at a given point on a curve is what the derivative tells us.
Imagine a curve on a graph. The tangent slope at a point gives you insight into the curve's steepness at precisely that point.
To approximate this slope, we often use a small piece of the curve and apply different methods, such as drawing lines through points positioned very close to each other. This method provides an estimation of the tangent slope when finding the exact derivative is complex.

Function Evaluation Basics

Function evaluation means plugging an input value into a function to get an output. Evaluating the function helps us find points on the graph of the function.
For instance, if we have a function \( f(x) = \frac{x^2 + x + 1}{x^2 + 2} \) and we want to know the value at \( x = 1 \), we substitute 1 into this function:

• This gives \( f(1) = \frac{1^2 + 1 + 1}{1^2 + 2} = 1 \).
• This process tells us that the point \( P = (1, 1) \) is on the graph of this function.

Understanding this basic step is crucial for more complex tasks like calculating derivatives as it establishes where lines like tangents are located compared to the curve.

Numerical Methods for Derivative Approximation

Sometimes, finding the exact derivative of a function is difficult, especially for complex functions. To overcome this, we use numerical methods to approximate the derivative.
One such method involves choosing two points close to the point of interest, called \( Q \) and \( R \). For example, by choosing \( Q = (0.9, f(0.9)) \) and \( R = (1.1, f(1.1)) \), and finding their slopes, we estimate the curve's steepness.
The formula used is:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \) which calculates the slope between two points.
• This approximated slope serves as a good estimation of \( f'(c) \), where \( c \) is our x-value of focus.

Numerical approximation is a practical approach in calculus, providing results when traditional calculus methods are too difficult to handle directly.