Let R be the set of all real numbers and f : R`rarr` R be given by f(x) = `3x^(2) + 1`. Then the set `f^(-1)` ([1, 6]) is

Options

A. `{-sqrt(5/3),0sqrt(5/3)}`
B. `[-sqrt(5/3),sqrt(5/3)]`
C. `{-sqrt(1/3),sqrt(1/3)}`
D. `(-sqrt(5/3),sqrt(5/3))`

Answer

To find the set \( f^{-1}([1, 6]) \) for the function \( f(x) = 3x^2 + 1 \), we will follow these steps:

### Step 1: Set up the equation
We want to find the values of \( x \) such that \( f(x) \) falls within the interval \([1, 6]\). This means we need to solve the inequalities:
\[ 1 \leq f(x) \leq 6 \]
Substituting \( f(x) \) into the inequality gives:
\[ 1 \leq 3x^2 + 1 \leq 6 \]

### Step 2: Simplify the inequalities
We can simplify the inequalities by subtracting 1 from all parts:
\[ 0 \leq 3x^2 \leq 5 \]

### Step 3: Divide by 3
Next, we divide the entire inequality by 3:
\[ 0 \leq x^2 \leq \frac{5}{3} \]

### Step 4: Solve for \( x \)
Taking the square root of the inequality gives us:
\[ 0 \leq x \leq \sqrt{\frac{5}{3}} \]
Since \( x^2 \) can also be negative, we also consider the negative root:
\[ -\sqrt{\frac{5}{3}} \leq x \leq 0 \]

### Step 5: Combine the results
Combining both results, we find:
\[ -\sqrt{\frac{5}{3}} \leq x \leq \sqrt{\frac{5}{3}} \]

### Final Result
Thus, the set \( f^{-1}([1, 6]) \) is:
\[ f^{-1}([1, 6]) = \left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right] \]