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To find the point(s) on the graph of the function \( f(x) = x^3 - 12x + 1 \) where the tangent line is horizontal, we need to follow these steps:

1. **Find the derivative** of the function. The derivative \( f'(x) \) represents the slope of the tangent line at any point \( x \). For a horizontal tangent line, the slope must be zero, so we solve \( f'(x) = 0 \).

2. **Solve \( f'(x) = 0 \) for \( x \)**. The solutions to this equation will give the \( x \)-coordinates where the function has a horizontal tangent line.

3. **Find the corresponding \( y \)-coordinates** by substituting the \( x \)-values back into the original function \( f(x) \).

Let's compute the derivative and solve this step-by-step.

### Step 1: Find the derivative of \( f(x) \)

The function is \( f(x) = x^3 - 12x + 1 \).

Using the power rule for derivatives:
\[
f'(x) = 3x^2 - 12
\]

### Step 2: Solve \( f'(x) = 0 \)

Set the derivative equal to zero:
\[
3x^2 - 12 = 0
\]

Solve for \( x \):
\[
3x^2 = 12
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]

### Step 3: Find the corresponding \( y \)-coordinates

Now that we have \( x = 2 \) and \( x = -2 \), we substitute these values into the original function \( f(x) \) to find the corresponding \( y \)-coordinates.

For \( x = 2 \):
\[
f(2) = (2)^3 - 12(2) + 1 = 8 - 24 + 1 = -15
\]

For \( x = -2 \):
\[
f(-2) = (-2)^3 - 12(-2) + 1 = -8 + 24 + 1 = 17
\]

### Final Answer

The points where the function has a horizontal tangent line are:
\[
(2, -15) \quad \text{and} \quad (-2, 17)
\]

Would you like further clarification or more details?

Here are 5 related questions:
1. What is the significance of the derivative when analyzing the behavior of a function?
2. How do you find the points of inflection for a cubic function like this one?
3. Why is setting the derivative equal to zero important for finding horizontal tangent lines?
4. What are critical points, and how do they relate to the function's increasing or decreasing behavior?
5. How would the behavior of the function change if the cubic term had a different coefficient?

**Tip:** Always check the second derivative if you need to determine whether a function is concave up or down at points where the tangent line is horizontal.

Find the point(s) on the graph of f(x) = x^3 - 12x + 1 for which the function has a horizontal tangent line.

Learn how to find the points on the graph of f(x) = x^3 - 12x + 1 where the tangent line is horizontal. This calculus problem involves taking the derivative of a cubic function and solving for critical points where f'(x) = 0.

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