https://file.aiquickdraw.com/mathbot/file/1726858936844tjgwnwpd.jpg

To solve the problem of finding the slope of the tangent line using the difference quotient formula \(\frac{f(x+h) - f(x)}{h}\), follow these steps:

### 1. **Function Given**:
\[ f(x) = \frac{11}{3x} \]

The point of interest is \( (1, \frac{11}{3}) \), meaning we will calculate the slope at \( x = 1 \).

### 2. **Difference Quotient**:
The slope of the tangent line is given by the derivative. Using the difference quotient:
\[
m = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}
\]

### 3. **Calculate \( f(1+h) \)**:
We substitute \( 1+h \) into the function \( f(x) \):
\[
f(1+h) = \frac{11}{3(1+h)} = \frac{11}{3+3h}
\]

### 4. **Calculate \( f(1) \)**:
From the given function \( f(x) \), when \( x = 1 \):
\[
f(1) = \frac{11}{3(1)} = \frac{11}{3}
\]

### 5. **Formulate the Difference Quotient**:
Now substitute these values into the difference quotient:
\[
m = \lim_{h \to 0} \frac{\frac{11}{3+3h} - \frac{11}{3}}{h}
\]

### 6. **Simplify the Expression**:
To simplify, find a common denominator:
\[
m = \lim_{h \to 0} \frac{11(3) - 11(3+3h)}{(3)(3+3h)h} = \lim_{h \to 0} \frac{33 - (33 + 33h)}{9h + 9h^2}
\]
\[
m = \lim_{h \to 0} \frac{33 - 33 - 33h}{9h + 9h^2} = \lim_{h \to 0} \frac{-33h}{9h + 9h^2}
\]

Cancel \( h \) from the numerator and denominator:
\[
m = \lim_{h \to 0} \frac{-33}{9 + 9h}
\]

As \( h \to 0 \), the denominator becomes 9:
\[
m = \frac{-33}{9} = -\frac{11}{3}
\]

### 7. **Equation of the Tangent Line**:
The slope is \( m = -\frac{11}{3} \). Now, use the point-slope form of a line \( y - y_1 = m(x - x_1) \) where \( (x_1, y_1) = (1, \frac{11}{3}) \):
\[
y - \frac{11}{3} = -\frac{11}{3}(x - 1)
\]

Simplify the equation:
\[
y - \frac{11}{3} = -\frac{11}{3}x + \frac{11}{3}
\]
\[
y = -\frac{11}{3}x + \frac{11}{3} + \frac{11}{3} = -\frac{11}{3}x + \frac{22}{3}
\]

Thus, the equation of the tangent line is:
\[
y = -\frac{11}{3}x + \frac{22}{3}
\]

---

Would you like further details or clarifications? Here are some related questions:

1. What is the geometric meaning of the slope of the tangent line?
2. How does the difference quotient relate to the derivative?
3. Can we find the derivative of the function using rules of differentiation directly?
4. What is the difference between a secant line and a tangent line?
5. How can the point-slope form be converted to slope-intercept form?

**Tip**: The slope of a tangent line represents the instantaneous rate of change of the function at a specific point.

Find the slope of the tangent line to the graph of the function at the given point: f(x) = 11 / (3x) at (1, 11/3)

In this problem, we calculate the slope of the tangent line to the curve f(x) = 11 / (3x) at the point (1, 11/3) using the derivative. Follow the step-by-step solution for finding the slope and the equation of the tangent line.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation