Let `f" [1,2] to (-oo,oo)` be given by `f(x)=(x^(4)+3x^(2)+1)/(x^(2)+1)` then find value of in `[f_(max)]" in "[-1,2]` where [.] is greatest integer function :

Answer

To solve the problem, we need to find the maximum value of the function \( f(x) = \frac{x^4 + 3x^2 + 1}{x^2 + 1} \) in the interval \([-1, 2]\) and then apply the greatest integer function to that maximum value.

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \frac{x^4 + 3x^2 + 1}{x^2 + 1} \]
We can simplify it:
\[ f(x) = \frac{(x^2 + 1)^2 + x^2}{x^2 + 1} = x^2 + 1 + \frac{x^2}{x^2 + 1} \]

### Step 2: Differentiate the function
Next, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} \left( x^2 + 1 + \frac{x^2}{x^2 + 1} \right) \]
Using the quotient rule for the last term:
\[ f'(x) = 2x + \frac{(x^2 + 1)(2x) - x^2(2x)}{(x^2 + 1)^2} \]
This simplifies to:
\[ f'(x) = 2x + \frac{2x}{x^2 + 1} \]

### Step 3: Set the derivative to zero
To find critical points, we set \( f'(x) = 0 \):
\[ 2x + \frac{2x}{x^2 + 1} = 0 \]
Factoring out \( 2x \):
\[ 2x \left( 1 + \frac{1}{x^2 + 1} \right) = 0 \]
This gives us \( x = 0 \) as a critical point.

### Step 4: Evaluate the function at critical points and endpoints
We need to evaluate \( f(x) \) at \( x = -1, 0, \) and \( 2 \):
1. For \( x = -1 \):
\[ f(-1) = \frac{(-1)^4 + 3(-1)^2 + 1}{(-1)^2 + 1} = \frac{1 + 3 + 1}{1 + 1} = \frac{5}{2} = 2.5 \]
2. For \( x = 0 \):
\[ f(0) = \frac{0^4 + 3(0)^2 + 1}{0^2 + 1} = \frac{1}{1} = 1 \]
3. For \( x = 2 \):
\[ f(2) = \frac{2^4 + 3(2)^2 + 1}{(2)^2 + 1} = \frac{16 + 12 + 1}{4 + 1} = \frac{29}{5} = 5.8 \]

### Step 5: Determine the maximum value
Now we compare the values:
- \( f(-1) = 2.5 \)
- \( f(0) = 1 \)
- \( f(2) = 5.8 \)

The maximum value in the interval \([-1, 2]\) is \( 5.8 \).

### Step 6: Apply the greatest integer function
Finally, we apply the greatest integer function:
\[ [f_{\text{max}}] = [5.8] = 5 \]

### Final Answer
Thus, the value of \( [f_{\text{max}}] \) in \([-1, 2]\) is \( 5 \).

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