The maximum value of
`f(x)=(x-1)^(1/2)(x-2),1lexle9`,is

Options

A. 14sqrt(2)
B. 15
C. 17
D. 18

Answer

To find the maximum value of the function \( f(x) = (x-1)^{1/2}(x-2) \) for \( 1 \leq x \leq 9 \), we will follow these steps:

### Step 1: Evaluate the function at the endpoints
We will first calculate the values of \( f(x) \) at the endpoints of the interval.

1. **Calculate \( f(1) \)**:
\[ f(1) = (1-1)^{1/2}(1-2) = 0 \cdot (-1) = 0 \]

2. **Calculate \( f(9) \)**:
\[ f(9) = (9-1)^{1/2}(9-2) = (8)^{1/2}(7) = 2\sqrt{2} \cdot 7 = 14\sqrt{2} \]

### Step 2: Find the critical points
Next, we will find the derivative of \( f(x) \) and set it to zero to find critical points.

1. **Differentiate \( f(x) \)** using the product rule:
\[ f'(x) = \frac{d}{dx}[(x-1)^{1/2}] \cdot (x-2) + (x-1)^{1/2} \cdot \frac{d}{dx}[x-2] \]
Using the chain rule, we have:
\[ \frac{d}{dx}[(x-1)^{1/2}] = \frac{1}{2}(x-1)^{-1/2} \cdot 1 = \frac{1}{2\sqrt{x-1}} \]
Therefore:
\[ f'(x) = \frac{1}{2\sqrt{x-1}}(x-2) + (x-1)^{1/2}(1) \]
Simplifying:
\[ f'(x) = \frac{x-2}{2\sqrt{x-1}} + \sqrt{x-1} \]

2. **Set \( f'(x) = 0 \)**:
\[ \frac{x-2}{2\sqrt{x-1}} + \sqrt{x-1} = 0 \]
Rearranging gives:
\[ \frac{x-2}{2\sqrt{x-1}} = -\sqrt{x-1} \]
Multiplying both sides by \( 2\sqrt{x-1} \):
\[ x-2 = -2(x-1) \]
Simplifying:
\[ x - 2 = -2x + 2 \implies 3x = 4 \implies x = \frac{4}{3} \]

### Step 3: Evaluate \( f(x) \) at the critical point
Now we will evaluate \( f(x) \) at the critical point \( x = \frac{4}{3} \).

1. **Calculate \( f\left(\frac{4}{3}\right) \)**:
\[ f\left(\frac{4}{3}\right) = \left(\frac{4}{3}-1\right)^{1/2}\left(\frac{4}{3}-2\right) \]
This simplifies to:
\[ f\left(\frac{4}{3}\right) = \left(\frac{1}{3}\right)^{1/2}\left(-\frac{2}{3}\right) = \frac{1}{\sqrt{3}} \cdot \left(-\frac{2}{3}\right) = -\frac{2}{3\sqrt{3}} \]
Since this value is negative, it cannot be the maximum.

### Step 4: Compare values
Now we compare the values obtained:

- \( f(1) = 0 \)
- \( f(9) = 14\sqrt{2} \)
- \( f\left(\frac{4}{3}\right) = -\frac{2}{3\sqrt{3}} \)

The maximum value occurs at \( x = 9 \):
\[ \text{Maximum value} = 14\sqrt{2} \]

### Conclusion
The maximum value of \( f(x) \) in the interval \( [1, 9] \) is \( 14\sqrt{2} \).

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