If a function `f:CtoC` is defined by `f(x)=3x^(2)-1`, where C is the set of complex numbers, then the pre-images of -28 are

Options

A. 3,-3
B. 3i,-3i
C. 3i only
D. `-3i` only

Answer

To find the pre-images of -28 for the function \( f: \mathbb{C} \to \mathbb{C} \) defined by \( f(x) = 3x^2 - 1 \), we follow these steps:

### Step 1: Set up the equation
We want to find the pre-images of -28, which means we need to solve the equation:
\[ f(x) = -28 \]
This translates to:
\[ 3x^2 - 1 = -28 \]

### Step 2: Rearrange the equation
Add 1 to both sides to isolate the term with \( x^2 \):
\[ 3x^2 = -28 + 1 \]
\[ 3x^2 = -27 \]

### Step 3: Divide by 3
Next, divide both sides by 3 to solve for \( x^2 \):
\[ x^2 = \frac{-27}{3} \]
\[ x^2 = -9 \]

### Step 4: Take the square root
Now, take the square root of both sides. Remember that taking the square root of a negative number involves the imaginary unit \( i \):
\[ x = \pm \sqrt{-9} \]
\[ x = \pm \sqrt{-1 \cdot 9} \]
\[ x = \pm 3i \]

### Step 5: Write the final answer
Thus, the pre-images of -28 are:
\[ x = 3i \quad \text{and} \quad x = -3i \]

### Summary
The pre-images of -28 under the function \( f(x) = 3x^2 - 1 \) are \( 3i \) and \( -3i \).

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