To find the absolute maximum and minimum values of the function $f(x) = x^2 - x - 4$ on the interval $[0, 4]$, follow these steps:

Step 1: Find the critical points

First, compute the derivative of the function $f(x)$:

$f'(x) = \frac{d}{dx}(x^2 - x - 4) = 2x - 1$

Set the derivative equal to zero to find the critical points:

$2x - 1 = 0$

Solving for $x$:

$x = \frac{1}{2}$

So, the critical point is $x = \frac{1}{2}$.

Step 2: Evaluate the function at the critical point and endpoints

Next, evaluate $f(x)$ at the critical point and the endpoints of the interval, $x = 0$ and $x = 4$.

1. $f(0) = (0)^2 - (0) - 4 = -4$
2. $f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} - 4 = \frac{1}{4} - \frac{1}{2} - 4 = -\frac{17}{4}$
3. $f(4) = (4)^2 - (4) - 4 = 16 - 4 - 4 = 8$

Step 3: Identify the absolute maximum and minimum

• At $x = 0$, $f(0) = -4$
• At $x = \frac{1}{2}$, $f\left(\frac{1}{2}\right) = -\frac{17}{4} \approx -4.25$
• At $x = 4$, $f(4) = 8$

Thus, the absolute minimum value is $f\left(\frac{1}{2}\right) = -\frac{17}{4}$, and the absolute maximum value is $f(4) = 8$.

Final Answer:

• Absolute minimum: $-\frac{17}{4}$
• Absolute maximum: $8$

Would you like more details or have any questions?

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Here are 5 related questions for further practice:

1. Find the critical points of the function $f(x) = x^2 - 6x + 9$ on the interval $[1, 5]$.
2. Determine the absolute maximum and minimum values of $f(x) = x^3 - 3x^2 + 2$ on $[0, 2]$.
3. Find the critical points and classify them as minima or maxima for $f(x) = 2x^3 - 9x^2 + 12x - 1$.
4. Evaluate the absolute extrema of $f(x) = \sin(x)$ on the interval $[0, \pi]$.
5. Solve for the absolute maximum and minimum values of $f(x) = e^{-x^2}$ on the interval $[-2, 2]$.

Tip: Always check both the critical points and the endpoints when determining the absolute extrema of a function on a closed interval.