To find the absolute maximum and minimum values of the function $f(x) = x^3 + 6x^2 + 1$ on the interval $[-5, 3]$, we need to follow these steps:

Step 1: Find the derivative of the function.

The first step is to find the first derivative of $f(x)$, as the critical points occur where the derivative equals zero or is undefined.

$f'(x) = \frac{d}{dx}(x^3 + 6x^2 + 1)$ Using basic differentiation rules: $f'(x) = 3x^2 + 12x$

Step 2: Solve $f'(x) = 0$ to find critical points.

Next, we solve for $x$ where $f'(x) = 0$:

$3x^2 + 12x = 0$

Factor the equation:

$3x(x + 4) = 0$

So, the solutions are:

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

Thus, the critical points are $x = 0$ and $x = -4$.

Step 3: Evaluate $f(x)$ at the critical points and endpoints.

Now, we evaluate the function at the critical points and at the endpoints of the interval, $x = -5$ and $x = 3$.

1. At $x = -5$:

$f(-5) = (-5)^3 + 6(-5)^2 + 1 = -125 + 6(25) + 1 = -125 + 150 + 1 = 26$

2. At $x = 3$:

$f(3) = (3)^3 + 6(3)^2 + 1 = 27 + 6(9) + 1 = 27 + 54 + 1 = 82$

3. At $x = 0$:

$f(0) = (0)^3 + 6(0)^2 + 1 = 1$

4. At $x = -4$:

$f(-4) = (-4)^3 + 6(-4)^2 + 1 = -64 + 6(16) + 1 = -64 + 96 + 1 = 33$

Step 4: Identify the absolute maximum and minimum.

Now we compare the function values at the critical points and endpoints:

• $f(-5) = 26$
• $f(3) = 82$
• $f(0) = 1$
• $f(-4) = 33$

The absolute maximum value of $f(x)$ on the interval $[-5, 3]$ is $\boxed{82}$, occurring at $x = 3$.

The absolute minimum value of $f(x)$ on the interval $[-5, 3]$ is $\boxed{1}$, occurring at $x = 0$.

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Let me know if you'd like further details on any part of this process or if you have any questions! Here are some related questions you might find helpful:

1. How do you determine critical points for other types of functions?
2. What role do the endpoints play in finding absolute maximum and minimum values?
3. How do you determine whether a critical point is a maximum, minimum, or a saddle point?
4. How do you apply the first and second derivative tests to classify critical points?
5. What is the significance of the second derivative in determining concavity and inflection points?

Tip: Always check both the critical points and the endpoints of a given interval when searching for absolute extrema.