4. On the closed interval $$ [0,5] $$, the function $$ f(x)=3-|x-1| $$ han (A) both an absolute maximum and an absolute minimum. (B) an absolute maximum but no absolute minimum. (C) no absolute maximum but an absolute minimum. (D) an absolute maximum and two absolute minima. 5. The critical numbers of the function

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To determine the nature of the function $$ f(x) = 3 - |x - 1| $$ on the closed interval $$[0, 5]$$, let's analyze its behavior:

1. **Understanding the Function:**
   - The function $$ f(x) = 3 - |x - 1| $$ represents a "V" shape with its vertex at $$ x = 1 $$.
   - The absolute value function creates a piecewise linear function:
     $$
     f(x) =
     \begin{cases}
       3 - (1 - x) = 2 + x & \text{if } x \leq 1 \
       3 - (x - 1) = 4 - x & \text{if } x > 1
     \end{cases}
     $$

2. **Evaluating Critical Points and Endpoints:**
   - **At $$ x = 1 $$:**
     $$
     f(1) = 3 - |1 - 1| = 3
     $$
     This is the vertex of the "V" and serves as a local (and absolute) maximum.
   
   - **At $$ x = 0 $$:**
     $$
     f(0) = 3 - |0 - 1| = 3 - 1 = 2
     $$
   
   - **At $$ x = 5 $$:**
     $$
     f(5) = 3 - |5 - 1| = 3 - 4 = -1
     $$
     This is the lowest point of the function on the interval, serving as the absolute minimum.

3. **Conclusion:**
   - **Absolute Maximum:** $$ f(1) = 3 $$
   - **Absolute Minimum:** $$ f(5) = -1 $$

Therefore, on the interval $$[0, 5]$$, the function $$ f(x) = 3 - |x - 1| $$ **has both an absolute maximum and an absolute minimum**.

**Answer:**  
**Problem 4 Answer:** (A) both an absolute maximum and an absolute minimum.
On the interval $$[0, 5]$$, the function $$ f(x) = 3 - |x - 1| $$ has both an absolute maximum and an absolute minimum. **Answer:** (A) both an absolute maximum and an absolute minimum.

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