Problem 71 - Calculus Single Variable

Question

A function \(f\) and four values \(c\) \(x_{1}, x_{2}\) and \(x_{3}\) are given. Using an appropriate viewing rectangle centered about the point \(P=(c, f(c)),\) graph \(f\) and the three secant lines passing through \(P\) that are determined by \(\left(x_{1}, f\left(x_{1}\right)\right),\left(x_{2}, f\left(x_{2}\right)\right),\) and \(\left(x_{3}, f\left(x_{3}\right)\right) .\) Use the secant line through \(\left(x_{3}, f\left(x_{3}\right)\right)\) to estimate \(f^{\prime}(c)\) $$ f(x)=x /(x+1), c=1, x_{1}=1.8, x_{2}=1.5, x_{3}=1.2 $$

Step-by-Step Solution

Answer

The estimated derivative \( f'(1) \) is approximately 0.227.

Step 1: Calculate Point P

Since the function given is \( f(x) = \frac{x}{x+1} \) and \( c = 1 \), we need to find the point \( P = (c, f(c)) \). To do this, substitute \( c = 1 \) into the function: \[ f(1) = \frac{1}{1+1} = \frac{1}{2} \]Thus, the point \( P = (1, \frac{1}{2}) \).

Step 2: Calculate Function Values

Calculate \( f(x_1) \), \( f(x_2) \), and \( f(x_3) \) using the function \( f(x) = \frac{x}{x+1} \):- \( f(1.8) = \frac{1.8}{2.8} = \frac{9}{14} \)- \( f(1.5) = \frac{1.5}{2.5} = \frac{3}{5} \)- \( f(1.2) = \frac{1.2}{2.2} = \frac{6}{11} \)

Step 3: Determine Secant Lines

The slope of a secant line through two points \((x_i, f(x_i))\) and \((c, f(c))\) is given by:\[ m = \frac{f(x_i) - f(c)}{x_i - c} \]Calculate the slopes for each set of points:- For \( x_1 = 1.8 \): \[ m = \frac{\frac{9}{14} - \frac{1}{2}}{1.8 - 1} = \frac{\frac{9}{14} - \frac{7}{14}}{0.8} = \frac{2}{11.2} \approx 0.179 \]- For \( x_2 = 1.5 \): \[ m = \frac{\frac{3}{5} - \frac{1}{2}}{1.5 - 1} = \frac{\frac{6}{10} - \frac{5}{10}}{0.5} = \frac{1}{5} = 0.2 \]- For \( x_3 = 1.2 \): \[ m = \frac{\frac{6}{11} - \frac{1}{2}}{1.2 - 1} = \frac{\frac{12}{22} - \frac{11}{22}}{0.2} = \frac{1}{4.4} \approx 0.227 \]

Step 4: Estimate Derivative at c

To estimate \( f'(c) \), use the slope of the secant line through point \((x_3, f(x_3))\), which is approximately \( 0.227 \). This is your estimate for \( f'(1) \).

Step 5: Graph the Function and Secant Lines

Graph the function \( f(x) = \frac{x}{x+1} \) centered at point \( P = (1, \frac{1}{2}) \). Also plot the three secant lines with the calculated slopes passing through \( P \) using the points:- \( x_1 = 1.8 \)- \( x_2 = 1.5 \)- \( x_3 = 1.2 \)

Key Concepts

Derivative Estimation

To understand derivative estimation, imagine it as finding the slope of the tangent line to a curve at a specific point. However, when calculating a derivative directly isn't feasible, we can use the slope of a secant line as an approximation. A secant line intersects two points on a curve, offering a 'best guess' of the slope at a specific point if those two points are very close together. Here's a simple breakdown:

• If you want to estimate the derivative at a point \( c \), look at a point close to \( c \).
• The slope of the secant line connecting these two points gives an approximate value of the derivative.
• As the distance between these points approaches zero, the secant line's slope more closely resembles the true derivative.

In our exercise, we calculated the slope using the point \((x_3, f(x_3))\), which gives an approximate derivative of the function \( f(x) = \frac{x}{x+1} \) at \( x=1 \), resulting in \( f'(1) \approx 0.227 \). This method is particularly useful when graphing software or analytical calculus tools aren't available.

Slope Calculation

The calculation of slope for secant lines is a critical tool in calculus. The slope of a line is essentially how steep it is, and for a secant, it can be found by dividing the difference in \( y \)-values by the difference in \( x \)-values. Here's how it works:

• For two points on a curve, say \((x_i, f(x_i))\) and \((c, f(c))\), the formula is \( m = \frac{f(x_i) - f(c)}{x_i - c} \).
• This gives you the average rate of change of the function between those two points.
• In the exercise's context, different \( x \) values are recorded: 1.8, 1.5, and 1.2, each compared to \( c=1 \).

Calculating each slope gives us a clearer idea of how \( f(x) \) changes on average. For instance, with \( x_3 = 1.2 \), the slope is approximately \( 0.227 \), providing insight into the behavior of the function on that interval.

Graphing Functions

Graphing helps visualize complex mathematical functions. When graphing functions like \( f(x) = \frac{x}{x+1} \), you should focus on critical points and behavior. Follow these steps to graph functions, including secant lines:

• Identify critical points, like \( P(1, \frac{1}{2}) \) in this exercise, and plot them accurately.
• Look for intercepts and asymptotes. For \( f(x) = \frac{x}{x+1} \), note that \( f(x) \) approaches \( 1 \) as \( x \) becomes very large, indicating a horizontal asymptote.
• Plot the function within a suitable viewing window, focusing on the behavior around point \( P \). Center your graph to get a clear view.
• Next, draw the secant lines using calculated slopes, such as \( 0.227 \) for \( x_3 \).

This combination of plotted points and lines enhances understanding, showing not only where the function lies but also its direction at different regions. Graphing these elements makes the theoretical concepts tangible and accessible, helping solidify learning.