To find the critical points of the function $f(x) = 4x^3 + 24x^2$, we follow these steps:

Step 1: Find the first derivative of $f(x)$

The critical points occur where the derivative is zero or undefined. Let's compute the first derivative:

$f'(x) = \frac{d}{dx}(4x^3 + 24x^2)$

Using the power rule:

$f'(x) = 12x^2 + 48x$

Step 2: Set the first derivative equal to zero

To find the critical points, set $f'(x) = 0$:

$12x^2 + 48x = 0$

Factor out $12x$:

$12x(x + 4) = 0$

This gives two solutions:

$x = 0 \quad \text{or} \quad x = -4$

Step 3: Verify whether these points are minima, maxima, or saddle points

To classify the critical points, we can use the second derivative test. First, we find the second derivative:

$f''(x) = \frac{d}{dx}(12x^2 + 48x) = 24x + 48$

Now, we evaluate $f''(x)$ at the critical points:

1. For $x = 0$:

$f''(0) = 24(0) + 48 = 48 \quad (\text{positive, so } x = 0 \text{ is a local minimum})$

2. For $x = -4$:

$f''(-4) = 24(-4) + 48 = -96 + 48 = -48 \quad (\text{negative, so } x = -4 \text{ is a local maximum})$

Conclusion:

• $x = 0$ is a local minimum.
• $x = -4$ is a local maximum.

Critical points:

The critical points are $x = 0$ and $x = -4$.

Would you like further details on this or explanations on how to interpret this result?

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Here are five related questions:

1. How can you find inflection points using the second derivative?
2. What is the significance of the second derivative being positive or negative?
3. How do you apply the first derivative test to classify critical points?
4. How does one determine concavity of a function from its second derivative?
5. Can a function have a critical point where the derivative does not exist?

Tip: When performing optimization problems, always double-check if the critical points lead to a local maximum, minimum, or a saddle point using either the first or second derivative test.