Let's break down the problem from the image.

The function given is: $f(x) = (x^2 - 4x - 12)^5$

We are asked to find the $x$-coordinates of all points where the tangent line to the curve is horizontal. To solve this:

Steps:

1. Differentiate $f(x)$ to find $f'(x)$.

   Using the chain rule, the derivative of $f(x) = (x^2 - 4x - 12)^5$ is: $f'(x) = 5(x^2 - 4x - 12)^4 \cdot (2x - 4)$ The first part $5(x^2 - 4x - 12)^4$ comes from the power rule, and the second part $(2x - 4)$ is the derivative of the inner function $x^2 - 4x - 12$.

2. Set $f'(x) = 0$ to find where the tangent line is horizontal. A tangent line is horizontal where the derivative equals zero.

   $5(x^2 - 4x - 12)^4 \cdot (2x - 4) = 0$

3. Solve the equation:

   • For the equation to be zero, either $5(x^2 - 4x - 12)^4 = 0$ or $(2x - 4) = 0$.
   • The term $5(x^2 - 4x - 12)^4 = 0$ has no real solutions because a fourth power cannot equal zero unless the expression inside is zero, and in this case, $x^2 - 4x - 12 = 0$ does not produce any solutions that make the whole term zero.
   • So, focus on solving $(2x - 4) = 0$.

4. Solve for $x$: $2x - 4 = 0$ $2x = 4$ $x = 2$

So, the $x$-coordinate where the tangent line is horizontal is $x = 2$.

Would you like further details on any step or have any questions?

Here are 5 related questions to expand on this topic:

1. What is the significance of a horizontal tangent line on a curve?
2. How do you apply the chain rule when differentiating composite functions like $f(x)$?
3. Can you have multiple points where the tangent line is horizontal? If so, how do you find them?
4. What would the second derivative tell us about the behavior of the function at $x = 2$?
5. How would you interpret the graph of $f(x)$ near the point $x = 2$?

Tip: When solving derivative problems, always check if the entire derivative expression could equal zero or only parts of it, as seen in cases involving products of functions.