Let's walk through the steps for solving the problem you've uploaded.

1. Find $f'(x)$:

Given:
$f(x) = x^3 - 7x^2$
To find $f'(x)$, apply the power rule:
$f'(x) = 3x^2 - 14x$
This part is correctly calculated and is shown as correct in the image.

2. Solve $f'(x) = 0$ for $x$:

We now solve the equation:
$3x^2 - 14x = 0$
Factor the equation:
$x(3x - 14) = 0$

The solutions are:
$x = 0 \quad \text{or} \quad 3x - 14 = 0$
$x = 0 \quad \text{or} \quad x = \frac{14}{3}$

So, the correct $x$-values are $x = 0$ and $x = \frac{14}{3}$.

3. Find the points where the tangent line is horizontal:

To find the points where the tangent line is horizontal, we substitute these $x$-values into the original function $f(x)$.

For $x = 0$:

Substitute $x = 0$ into $f(x)$: $f(0) = 0^3 - 7(0^2) = 0$
So the point is $(0, 0)$.

For $x = \frac{14}{3}$:

Substitute $x = \frac{14}{3}$ into $f(x)$: $f\left(\frac{14}{3}\right) = \left(\frac{14}{3}\right)^3 - 7\left(\frac{14}{3}\right)^2$

Let's compute this value.

For $x = \frac{14}{3}$, the corresponding value of $f(x)$ is approximately $-50.815$.

Thus, the points where the tangent line is horizontal are:

• Smaller $x$-value: $(0, 0)$
• Larger $x$-value: $\left( \frac{14}{3}, -50.815 \right)$

Would you like more details or explanations about any steps? Here are some related questions to deepen your understanding:

1. What is the significance of finding where the tangent line is horizontal?
2. Can there be more than two points where the derivative is zero in a cubic function?
3. How can the critical points of a function help identify its maxima and minima?
4. How would you determine if these points are local maxima, minima, or saddle points?
5. What changes would occur in the function if the coefficients in the original equation were different?

Tip: Whenever you solve equations involving derivatives, always verify if the solutions represent maximum, minimum, or inflection points by checking the second derivative or analyzing the function graph.