Maximum value of the function `f(x) =(x)/(8)+(2)/(x)` on the interval [1,6] is

Options

A. 1
B. `(9)/(8)`
C. `(13)/(12)`
D. `(17)/(8)`

Answer

To find the maximum value of the function \( f(x) = \frac{x}{8} + \frac{2}{x} \) on the interval \([1, 6]\), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{x}{8}\right) + \frac{d}{dx}\left(\frac{2}{x}\right) \]
The derivative of \( \frac{x}{8} \) is \( \frac{1}{8} \) and the derivative of \( \frac{2}{x} \) can be computed using the power rule:
\[ \frac{d}{dx}\left(\frac{2}{x}\right) = \frac{d}{dx}(2x^{-1}) = -2x^{-2} = -\frac{2}{x^2} \]
Thus, the derivative is:
\[ f'(x) = \frac{1}{8} - \frac{2}{x^2} \]

### Step 2: Set the derivative to zero to find critical points
To find the critical points, we set the derivative equal to zero:
\[ \frac{1}{8} - \frac{2}{x^2} = 0 \]
Rearranging gives:
\[ \frac{2}{x^2} = \frac{1}{8} \]
Cross-multiplying yields:
\[ 2 \cdot 8 = x^2 \implies x^2 = 16 \implies x = 4 \quad (\text{since } x \text{ must be positive}) \]

### Step 3: Evaluate the function at critical points and endpoints
Now we need to evaluate \( f(x) \) at the critical point \( x = 4 \) and at the endpoints of the interval \( x = 1 \) and \( x = 6 \).

1. **At \( x = 1 \)**:
\[ f(1) = \frac{1}{8} + \frac{2}{1} = \frac{1}{8} + 2 = \frac{1}{8} + \frac{16}{8} = \frac{17}{8} \]

2. **At \( x = 4 \)**:
\[ f(4) = \frac{4}{8} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 \]

3. **At \( x = 6 \)**:
\[ f(6) = \frac{6}{8} + \frac{2}{6} = \frac{3}{4} + \frac{1}{3} \]
To add these fractions, we find a common denominator (which is 12):
\[ f(6) = \frac{3 \times 3}{4 \times 3} + \frac{1 \times 4}{3 \times 4} = \frac{9}{12} + \frac{4}{12} = \frac{13}{12} \]

### Step 4: Compare the values
Now we compare the values:
- \( f(1) = \frac{17}{8} \approx 2.125 \)
- \( f(4) = 1 \)
- \( f(6) = \frac{13}{12} \approx 1.0833 \)

### Conclusion
The maximum value of \( f(x) \) on the interval \([1, 6]\) is:
\[ \boxed{\frac{17}{8}} \]