Find the absolute maximum and absolute minimum value of `f(x) =(x-2) sqrt(x-1)` in [1,9]

Answer

To find the absolute maximum and absolute minimum values of the function \( f(x) = (x-2) \sqrt{x-1} \) on the interval \([1, 9]\), we will follow these steps:

### Step 1: Evaluate the function at the endpoints of the interval
First, we will calculate \( f(1) \) and \( f(9) \).

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

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

### Step 2: Find critical points by differentiating the function
Next, we need to find the derivative \( f'(x) \) and set it to zero to find critical points.

1. **Differentiate \( f(x) \)**:
Using the product rule:
\[ f'(x) = \frac{d}{dx}[(x-2) \sqrt{x-1}] = (x-2) \cdot \frac{1}{2\sqrt{x-1}} + \sqrt{x-1} \cdot 1 \]
Simplifying:
\[ f'(x) = \frac{(x-2)}{2\sqrt{x-1}} + \sqrt{x-1} \]
Combine the terms:
\[ f'(x) = \frac{(x-2) + 2(x-1)}{2\sqrt{x-1}} = \frac{3x - 4}{2\sqrt{x-1}} \]

2. **Set \( f'(x) = 0 \)**:
\[ \frac{3x - 4}{2\sqrt{x-1}} = 0 \implies 3x - 4 = 0 \implies x = \frac{4}{3} \]

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

1. **Calculate \( f\left(\frac{4}{3}\right) \)**:
\[ f\left(\frac{4}{3}\right) = \left(\frac{4}{3} - 2\right) \sqrt{\frac{4}{3} - 1} = \left(-\frac{2}{3}\right) \sqrt{\frac{1}{3}} = -\frac{2}{3} \cdot \frac{1}{\sqrt{3}} = -\frac{2}{3\sqrt{3}} = -\frac{2\sqrt{3}}{9} \]

### Step 4: Compare values to find absolute maximum and minimum
Now we compare the values obtained:

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

### Conclusion
- The absolute maximum value is \( f(9) = 14\sqrt{2} \).
- The absolute minimum value is \( f\left(\frac{4}{3}\right) = -\frac{2\sqrt{3}}{9} \).