Math Problem Statement

For each value of x given in the table below, find the slope of the secant line passing through (0, 8) and (x, f(x)). Do not round intermediate computations, and round your answers to 4 decimal places if necessary.

Solution

This exercise involves estimating the slope of the tangent line at the point (0,8)(0, 8) on the graph of the function f(x)=16x+2f(x) = \frac{16}{x+2}, by using the slopes of several secant lines.

To find the slope of a secant line passing through the points (0,8)(0, 8) and (x,f(x))(x, f(x)), we use the formula for the slope between two points:

slope of secant line=f(x)−f(0)x−0\text{slope of secant line} = \frac{f(x) - f(0)}{x - 0}

Given that f(0)=8f(0) = 8 (since f(x)=16x+2f(x) = \frac{16}{x+2}, and f(0)=160+2=8f(0) = \frac{16}{0+2} = 8):

slope of secant line=f(x)−8x\text{slope of secant line} = \frac{f(x) - 8}{x}

Now, for each xx value provided in the table, we substitute into f(x)=16x+2f(x) = \frac{16}{x+2} to compute the slope.

Let me calculate the slopes for each xx value in the table.Here are the slopes of the secant lines for each xx value in the table:

\hline \text{Value of } x & \text{Slope} \\ \hline -0.2 & -4.4444 \\ -0.1 & -4.2105 \\ -0.01 & -4.0201 \\ -0.001 & -4.0020 \\ 0.001 & -3.9980 \\ 0.01 & -3.9801 \\ 0.1 & -3.8095 \\ 0.2 & -3.6364 \\ \hline \end{array}$$ These slopes give an estimate of the slope of the tangent line at $$(0, 8)$$, which appears to be approaching a value near $$-4$$. Would you like more details or have any further questions? Here are 5 related questions to deepen your understanding: 1. How does the slope of the secant line change as $$x$$ approaches zero? 2. What is the mathematical definition of a tangent line in calculus? 3. Can you explain why the slopes approach a specific value as $$x \to 0$$? 4. How would you use limits to find the exact slope of the tangent line at $$(0, 8)$$? 5. What is the derivative of $$f(x) = \frac{16}{x+2}$$, and how does it relate to these secant slopes? **Tip**: The slope of the tangent line at a point can be found using the derivative of the function at that point!

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Limits
Secant Lines
Tangent Lines

Formulas

Slope of secant line = (f(x) - f(0)) / (x - 0)
f(x) = 16 / (x + 2)

Theorems

Definition of a Derivative
Secant and Tangent Line Approximations

Suitable Grade Level

Grades 10-12

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