Let's solve the problems step by step for both functions, starting with:

Problem 45: $f(x) = 12 + 4x - x^2$ on the interval $[0,5]$

1. Find the first derivative of the function to identify critical points: $f'(x) = \frac{d}{dx}(12 + 4x - x^2) = 4 - 2x$

2. Set the derivative equal to zero to find critical points: $4 - 2x = 0$ Solving for $x$: $x = 2$ So, there is a critical point at $x = 2$.

3. Evaluate the function at the endpoints of the interval and the critical point: $f(0) = 12 + 4(0) - (0)^2 = 12$ $f(5) = 12 + 4(5) - (5)^2 = 12 + 20 - 25 = 7$ $f(2) = 12 + 4(2) - (2)^2 = 12 + 8 - 4 = 16$

4. Determine the absolute maximum and minimum:

   • At $x = 0$, $f(0) = 12$
   • At $x = 5$, $f(5) = 7$
   • At $x = 2$, $f(2) = 16$

   The absolute maximum value is $f(2) = 16$, and the absolute minimum value is $f(5) = 7$.

Problem 47: $f(x) = 2x^3 - 3x^2 - 12x + 1$ on the interval $[-2, 3]$

1. Find the first derivative of the function: $f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x + 1) = 6x^2 - 6x - 12$

2. Set the derivative equal to zero to find critical points: $6x^2 - 6x - 12 = 0$ Divide through by 6: $x^2 - x - 2 = 0$ Factor the quadratic equation: $(x - 2)(x + 1) = 0$ So, the critical points are $x = 2$ and $x = -1$.

3. Evaluate the function at the endpoints and the critical points: $f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) + 1 = -16 - 12 + 24 + 1 = -3$ $f(3) = 2(3)^3 - 3(3)^2 - 12(3) + 1 = 54 - 27 - 36 + 1 = -8$ $f(2) = 2(2)^3 - 3(2)^2 - 12(2) + 1 = 16 - 12 - 24 + 1 = -19$ $f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) + 1 = -2 - 3 + 12 + 1 = 8$

4. Determine the absolute maximum and minimum:

   • At $x = -2$, $f(-2) = -3$
   • At $x = 3$, $f(3) = -8$
   • At $x = 2$, $f(2) = -19$
   • At $x = -1$, $f(-1) = 8$

   The absolute maximum value is $f(-1) = 8$, and the absolute minimum value is $f(2) = -19$.

Summary of Results:

• For problem 45, $f(x) = 12 + 4x - x^2$ on $[0,5]$:

  • Absolute maximum value: $16$ at $x = 2$
  • Absolute minimum value: $7$ at $x = 5$

• For problem 47, $f(x) = 2x^3 - 3x^2 - 12x + 1$ on $[-2,3]$:

  • Absolute maximum value: $8$ at $x = -1$
  • Absolute minimum value: $-19$ at $x = 2$

Would you like me to go into further detail or have any questions about the steps?

Here are some follow-up questions that may help:

1. How do you find critical points for a higher-degree polynomial function?
2. Can the first derivative always give you all the critical points? What if a point is at the endpoint?
3. How would you determine if a critical point is a maximum or minimum using the second derivative test?
4. What is the importance of evaluating the function at the endpoints of an interval?
5. How does the graph of a function help in visualizing maximum and minimum values?

Tip: When finding the absolute maximum or minimum on a closed interval, always remember to check both the critical points and the endpoints of the interval.