https://file.aiquickdraw.com/mathbot/file/1727056035451uthcal6k.jpg

The problem in the image provides a function \( f(x) = \ln(x^2 - 6x + k) \), and the questions are:

1. **Find \( k \in \mathbb{R} \)** so that the slope of the tangent line to the graph of \( f(x) \) at \( x_0 = 4 \) is equal to 2.
2. **For the value of \( k \) found in the previous part,** explicitly find the equation of the tangent line to the graph of \( f(x) \) at \( x_0 = 4 \).

### Step 1: Finding \( k \)
To solve for \( k \), we need to calculate the derivative of \( f(x) \), which gives us the slope of the tangent line at any point \( x_0 \).

1. The function is \( f(x) = \ln(x^2 - 6x + k) \).
2. The derivative of \( f(x) \) is:
   \[
   f'(x) = \frac{d}{dx} \ln(x^2 - 6x + k) = \frac{1}{x^2 - 6x + k} \cdot (2x - 6)
   \]
   So:
   \[
   f'(x) = \frac{2x - 6}{x^2 - 6x + k}
   \]

3. At \( x_0 = 4 \), we set the slope equal to 2:
   \[
   f'(4) = 2
   \]
   Substituting \( x = 4 \) into the derivative:
   \[
   f'(4) = \frac{2(4) - 6}{4^2 - 6(4) + k} = \frac{8 - 6}{16 - 24 + k} = \frac{2}{-8 + k}
   \]
   We now solve:
   \[
   \frac{2}{-8 + k} = 2
   \]
   Multiplying both sides by \( -8 + k \):
   \[
   2 = 2(-8 + k)
   \]
   Simplifying:
   \[
   2 = -16 + 2k
   \]
   Solving for \( k \):
   \[
   18 = 2k \quad \Rightarrow \quad k = 9
   \]

### Step 2: Finding the equation of the tangent line
Now that we know \( k = 9 \), we can find the equation of the tangent line at \( x_0 = 4 \).

1. The point of tangency is \( (4, f(4)) \). To find \( f(4) \):
   \[
   f(4) = \ln(4^2 - 6(4) + 9) = \ln(16 - 24 + 9) = \ln(1) = 0
   \]
   So the point is \( (4, 0) \).

2. The slope is given as 2, so the equation of the tangent line is:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting \( m = 2 \), \( x_1 = 4 \), and \( y_1 = 0 \):
   \[
   y - 0 = 2(x - 4)
   \]
   Simplifying:
   \[
   y = 2(x - 4) = 2x - 8
   \]

Thus, the equation of the tangent line is \( y = 2x - 8 \).

---

Would you like more details or explanations about any step?

### Related Questions:
1. How do we find the derivative of a natural logarithmic function?
2. What is the geometric interpretation of the tangent line to a curve?
3. How do we calculate the value of \( f(x) \) for any given \( x \)?
4. Can the value of \( k \) be negative? Why or why not?
5. What happens if the slope condition changes to another value, e.g., 3?

### Tip:
When solving for tangents, remember that the derivative at a point gives the slope of the tangent line at that point.

Find k ∈ ℝ such that the slope of the tangent line to the graph of f(x) = ln(x^2 - 6x + k) at x_0 = 4 is equal to 2. For the value of k found, find the explicit equation of the tangent line at x_0 = 4.

Solve for k such that the slope of the tangent line to f(x) = ln(x^2 - 6x + k) at x_0 = 4 is equal to 2. Learn how to find the tangent line's equation at that point. Suitable for Grades 10-12, covering key calculus and algebra concepts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation