Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=x^(2)+16/x,x in [1, 3]`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = x^2 + \frac{16}{x} \) on the interval \([1, 3]\), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function to find the critical points.

\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}\left(\frac{16}{x}\right) \]

Calculating the derivatives:
- The derivative of \( x^2 \) is \( 2x \).
- The derivative of \( \frac{16}{x} \) can be rewritten as \( 16x^{-1} \), and its derivative is \( -\frac{16}{x^2} \).

Thus, we have:

\[ f'(x) = 2x - \frac{16}{x^2} \]

### Step 2: Set the derivative to zero to find critical points
We set the derivative equal to zero to find the critical points:

\[ 2x - \frac{16}{x^2} = 0 \]

Multiplying through by \( x^2 \) to eliminate the fraction gives:

\[ 2x^3 - 16 = 0 \]

Rearranging this, we find:

\[ 2x^3 = 16 \quad \Rightarrow \quad x^3 = 8 \quad \Rightarrow \quad x = 2 \]

### Step 3: Evaluate the function at critical points and endpoints
Now we will evaluate the function \( f(x) \) at the critical point \( x = 2 \) and at the endpoints of the interval \( x = 1 \) and \( x = 3 \).

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

2. **At \( x = 2 \)**:
\[ f(2) = 2^2 + \frac{16}{2} = 4 + 8 = 12 \]

3. **At \( x = 3 \)**:
\[ f(3) = 3^2 + \frac{16}{3} = 9 + \frac{16}{3} = 9 + 5.33 \approx 14.33 \]

### Step 4: Determine the absolute maximum and minimum
Now we compare the values obtained:

- \( f(1) = 17 \)
- \( f(2) = 12 \)
- \( f(3) \approx 14.33 \)

The absolute maximum value is \( 17 \) at \( x = 1 \), and the absolute minimum value is \( 12 \) at \( x = 2 \).

### Final Answer:
- **Absolute Maximum**: \( 17 \) at \( x = 1 \)
- **Absolute Minimum**: \( 12 \) at \( x = 2 \)

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