For what values of x
  does the function f(x)= 12x - x^3 have a horizontal tangent line

To find the values of \( x \) where the function \( f(x) = 12x - x^3 \) has a horizontal tangent line, we need to determine where the derivative \( f'(x) \) equals zero.

### Step 1: Find the derivative
The given function is:
\[
f(x) = 12x - x^3
\]
The derivative of \( f(x) \) with respect to \( x \) is:
\[
f'(x) = 12 - 3x^2
\]

### Step 2: Set the derivative equal to zero
To find where the tangent line is horizontal, set \( f'(x) = 0 \):
\[
12 - 3x^2 = 0
\]
Solving for \( x \):
\[
3x^2 = 12
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]

### Conclusion
The function \( f(x) = 12x - x^3 \) has horizontal tangent lines at \( x = 2 \) and \( x = -2 \).

Would you like more details or have any questions?

### Relative Questions
1. What is the meaning of a horizontal tangent line in terms of a function's graph?
2. How do you determine whether a critical point is a maximum, minimum, or neither?
3. How does the shape of the graph \( f(x) = 12x - x^3 \) look near the points \( x = 2 \) and \( x = -2 \)?
4. What role does the second derivative play in analyzing concavity around these points?
5. Can you find other functions with horizontal tangent lines at similar intervals?

### Tip
The derivative \( f'(x) = 0 \) represents points where the slope is zero, indicating possible maxima, minima, or points of inflection.

For what values of x does the function f(x) = 12x - x^3 have a horizontal tangent line?

Discover how to find the points where the function f(x) = 12x - x^3 has horizontal tangent lines by solving for when its derivative equals zero. This calculus problem involves understanding derivatives and critical points, suitable for students in Grades 11-12.

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