Given 𝑓(𝑥)=4/(3𝑥−1)

, find the slope of the secant line to 𝑓(𝑥) between 𝑥1=−5 and 𝑥2=8. 

To find the slope of the secant line between two points on a curve, we use the formula for the slope of a secant line between two points \((x_1, f(x_1))\) and \((x_2, f(x_2))\):

\[
\text{slope of secant line} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}
\]

Given the function:

\[
f(x) = \frac{4}{3x - 1}
\]

### Step 1: Calculate \( f(x_1) \) and \( f(x_2) \)

1. \( f(x_1) = f(-5) \):

\[
f(-5) = \frac{4}{3(-5) - 1} = \frac{4}{-15 - 1} = \frac{4}{-16} = -\frac{1}{4}
\]

2. \( f(x_2) = f(8) \):

\[
f(8) = \frac{4}{3(8) - 1} = \frac{4}{24 - 1} = \frac{4}{23}
\]

### Step 2: Use the secant line formula

\[
\text{slope} = \frac{f(8) - f(-5)}{8 - (-5)} = \frac{\frac{4}{23} - \left(-\frac{1}{4}\right)}{8 + 5}
\]

### Step 3: Simplify

First, find a common denominator for \(\frac{4}{23}\) and \(-\frac{1}{4}\):

\[
\frac{4}{23} + \frac{1}{4} = \frac{4 \cdot 4 + 1 \cdot 23}{23 \cdot 4} = \frac{16 + 23}{92} = \frac{39}{92}
\]

Now, the slope becomes:

\[
\text{slope} = \frac{\frac{39}{92}}{13} = \frac{39}{92 \cdot 13} = \frac{39}{1196}
\]

Thus, the slope of the secant line is:

\[
\boxed{\frac{39}{1196}}
\]

Would you like to go over any details or have any further questions?

Here are 5 related questions to explore:
1. How do you find the equation of a secant line?
2. What is the difference between a secant line and a tangent line?
3. How can you generalize the formula for the slope of a secant line?
4. How would you find the slope of the tangent line to this function at a point?
5. How does the secant line slope behave as \(x_1\) approaches \(x_2\)?

**Tip:** Secant lines provide an average rate of change, while tangent lines give the instantaneous rate of change.

Given f(x) = 4/(3x−1), find the slope of the secant line to f(x) between x1 = −5 and x2 = 8.

Discover how to calculate the slope of the secant line for the function f(x) = 4/(3x−1) between the points x1 = −5 and x2 = 8. This problem involves key algebraic and calculus concepts, suitable for students in Grades 11-12.

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