Find f(x) when it is given by
`f(x)=max{x^(3),x^(2),(1)/(64)},AA x in [0, oo).`

Answer

To find the function \( f(x) = \max\{x^3, x^2, \frac{1}{64}\} \) for \( x \in [0, \infty) \), we will analyze the three functions involved and determine which one is the maximum in different intervals.

### Step 1: Analyze the functions
We have three functions:
1. \( f_1(x) = x^3 \)
2. \( f_2(x) = x^2 \)
3. \( f_3(x) = \frac{1}{64} \)

### Step 2: Find intersection points
We need to find the points where these functions intersect, as these points will help us determine the intervals where each function is maximum.

1. **Intersection of \( x^2 \) and \( \frac{1}{64} \)**:
\[ x^2 = \frac{1}{64} \]
Taking the square root:
\[ x = \frac{1}{8} \]

2. **Intersection of \( x^3 \) and \( \frac{1}{64} \)**:
\[ x^3 = \frac{1}{64} \]
Taking the cube root:
\[ x = \frac{1}{4} \]

3. **Intersection of \( x^2 \) and \( x^3 \)**:
\[ x^2 = x^3 \implies x^2(1 - x) = 0 \]
This gives us \( x = 0 \) or \( x = 1 \).

### Step 3: Determine intervals
The critical points we found are \( x = 0, \frac{1}{8}, \frac{1}{4}, 1 \). We will evaluate the maximum function in the intervals defined by these points:
1. \( [0, \frac{1}{8}) \)
2. \( [\frac{1}{8}, \frac{1}{4}) \)
3. \( [\frac{1}{4}, 1) \)
4. \( [1, \infty) \)

### Step 4: Evaluate each interval
1. **For \( x \in [0, \frac{1}{8}) \)**:
- \( f_1(x) = x^3 \) is less than \( \frac{1}{64} \) and \( f_2(x) = x^2 \) is also less than \( \frac{1}{64} \).
- Therefore, \( f(x) = \frac{1}{64} \).

2. **For \( x \in [\frac{1}{8}, \frac{1}{4}) \)**:
- At \( x = \frac{1}{8} \), \( f_2(x) = \frac{1}{64} \) and \( f_1(x) = \frac{1}{512} \).
- \( f_2(x) \) is greater than \( f_1(x) \) and equal to \( f_3(x) \).
- Therefore, \( f(x) = x^2 \).

3. **For \( x \in [\frac{1}{4}, 1) \)**:
- At \( x = \frac{1}{4} \), \( f_1(x) = \frac{1}{64} \) and \( f_2(x) = \frac{1}{16} \).
- \( f_2(x) \) is greater than \( f_3(x) \) and \( f_1(x) \).
- Therefore, \( f(x) = x^2 \).

4. **For \( x \in [1, \infty) \)**:
- Both \( f_1(x) = x^3 \) and \( f_2(x) = x^2 \) are greater than \( \frac{1}{64} \).
- Since \( x^3 \) grows faster than \( x^2 \), \( f(x) = x^3 \).

### Final Answer
Putting it all together, we have:
\[ f(x) = \begin{cases} \frac{1}{64} & \text{if } x \in [0, \frac{1}{8}) \\ x^2 & \text{if } x \in [\frac{1}{8}, 1) \\ x^3 & \text{if } x \in [1, \infty) \end{cases} \]