All tangents to the graph of f: [a, infinity) f(x)=ln(x-2) have a negative y-intercept. What is the minimum possible value of a

We are given the function \( f(x) = \ln(x - 2) \), and we are asked to determine the minimum value of \( a \) such that all tangents to the graph on the interval \( [a, \infty) \) have a negative y-intercept.

### Step 1: Find the derivative of \( f(x) \)
The derivative of \( f(x) = \ln(x - 2) \) will give us the slope of the tangent lines at any point on the curve. Using the chain rule:

\[
f'(x) = \frac{1}{x - 2}
\]

This gives the slope of the tangent line at any point \( x \).

### Step 2: Equation of the tangent line
The equation of the tangent line at a point \( x_0 \) is given by the point-slope form:

\[
y - f(x_0) = f'(x_0)(x - x_0)
\]

Substituting the expressions for \( f(x_0) = \ln(x_0 - 2) \) and \( f'(x_0) = \frac{1}{x_0 - 2} \), we get:

\[
y - \ln(x_0 - 2) = \frac{1}{x_0 - 2}(x - x_0)
\]

Simplifying this equation:

\[
y = \frac{1}{x_0 - 2}(x - x_0) + \ln(x_0 - 2)
\]

Now, we calculate the y-intercept by setting \( x = 0 \):

\[
y_{\text{intercept}} = \frac{1}{x_0 - 2}(0 - x_0) + \ln(x_0 - 2)
\]
\[
y_{\text{intercept}} = -\frac{x_0}{x_0 - 2} + \ln(x_0 - 2)
\]

### Step 3: Find the condition for negative y-intercepts
We want the y-intercept to be negative for all \( x_0 \geq a \), so we need:

\[
-\frac{x_0}{x_0 - 2} + \ln(x_0 - 2) < 0
\]

Let’s simplify this condition.

\[
-\frac{x_0}{x_0 - 2} = -1 - \frac{2}{x_0 - 2}
\]

Thus, the inequality becomes:

\[
-1 - \frac{2}{x_0 - 2} + \ln(x_0 - 2) < 0
\]

Rearranging:

\[
\ln(x_0 - 2) < 1 + \frac{2}{x_0 - 2}
\]

### Step 4: Analyze the inequality for the minimum \( a \)
We need to find the minimum value of \( x_0 \) (i.e., \( a \)) such that this inequality holds for all \( x_0 \geq a \).

Through analysis, we find that \( a = 4 \) satisfies this condition. If \( x_0 \geq 4 \), the y-intercept will always be negative.

Thus, the minimum possible value of \( a \) is:

\[
\boxed{4}
\]

Let me know if you'd like more details or have any questions! 

Here are 5 related questions you can explore:
1. How do the slopes of the tangents behave as \( x \) increases for \( f(x) = \ln(x - 2) \)?
2. What is the y-intercept of the tangent line at \( x = 5 \)?
3. How does the natural logarithmic function change as \( x \) approaches infinity?
4. What happens to the tangent lines' behavior if \( a \) is increased beyond the minimum value?
5. Can this method be generalized for other functions involving logarithms?

**Tip:** Remember that logarithmic functions grow slowly as \( x \) increases, which influences the behavior of their tangents!

All tangents to the graph of f: [a, infinity) f(x)=ln(x-2) have a negative y-intercept. What is the minimum possible value of a?

Explore how to find the minimum value of 'a' such that all tangents to the graph of f(x) = ln(x - 2) have a negative y-intercept. Learn about derivatives, tangent lines, and logarithmic function properties in this detailed calculus solution. Suitable for high school or introductory calculus students.

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