To find the absolute minimum value of the function $f(x) = x^4 + 8x^3 + 18x^2 - 6$ on the closed interval $[-4, 1]$, we need to follow these steps:

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

The first derivative will help us locate the critical points of the function, which are potential candidates for local minima or maxima.

$f'(x) = \frac{d}{dx}(x^4 + 8x^3 + 18x^2 - 6)$

Using standard differentiation rules:

$f'(x) = 4x^3 + 24x^2 + 36x$

Step 2: Solve for critical points

We set the first derivative equal to zero to find the critical points.

$4x^3 + 24x^2 + 36x = 0$

Factor out the common term $4x$:

$4x(x^2 + 6x + 9) = 0$

We now solve for $x$. First, solve for $x = 0$.

Next, solve the quadratic equation:

$x^2 + 6x + 9 = 0$

This is a perfect square:

$(x + 3)^2 = 0$

Thus, $x = -3$ is a repeated root.

So, the critical points are $x = 0$ and $x = -3$.

Step 3: Evaluate the function at the critical points and endpoints of the interval

We now evaluate $f(x)$ at the critical points and the endpoints of the interval $[-4, 1]$.

• At $x = -4$:

$f(-4) = (-4)^4 + 8(-4)^3 + 18(-4)^2 - 6$ $= 256 - 512 + 288 - 6$ $= 26$

• At $x = -3$:

$f(-3) = (-3)^4 + 8(-3)^3 + 18(-3)^2 - 6$ $= 81 - 216 + 162 - 6$ $= 21$

• At $x = 0$:

$f(0) = 0^4 + 8(0)^3 + 18(0)^2 - 6$ $= -6$

• At $x = 1$:

$f(1) = 1^4 + 8(1)^3 + 18(1)^2 - 6$ $= 1 + 8 + 18 - 6$ $= 21$

Step 4: Determine the absolute minimum

From the evaluations, we have the following values:

• $f(-4) = 26$
• $f(-3) = 21$
• $f(0) = -6$
• $f(1) = 21$

The absolute minimum value of $f(x)$ on the interval $[-4, 1]$ is $\boxed{-6}$, which occurs at $x = 0$.

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Would you like more details on any of the steps? Here are some related questions that might help you deepen your understanding:

1. How do you use the second derivative test to classify critical points?
2. What is the significance of the endpoints in finding absolute extrema on a closed interval?
3. How can you determine whether a function has a local minimum or maximum at a given critical point?
4. What are the properties of polynomials of degree 4 that influence their graph and extrema?
5. How does the concept of concavity help identify local minima and maxima?

Tip: Always remember to check both the critical points and the endpoints of the interval when looking for absolute extrema on a closed interval!