The domain for which the functions defined by `f(x)=3x^(2)-1 and g(x)=3+x` are equal to :

Options

A. `[-1,4/3]`
B. `[1,4/3]`
C. `[-1,-4/3]`
D. `[-2,-4/3]`

Answer

To find the domain for which the functions \( f(x) = 3x^2 - 1 \) and \( g(x) = 3 + x \) are equal, we need to set the two functions equal to each other and solve for \( x \).

### Step-by-Step Solution:

1. **Set the functions equal to each other:**
\[ f(x) = g(x) \]
This gives us:
\[ 3x^2 - 1 = 3 + x \]

2. **Rearrange the equation:**
Move all terms to one side of the equation:
\[ 3x^2 - x - 1 - 3 = 0 \]
Simplifying this, we get:
\[ 3x^2 - x - 4 = 0 \]

3. **Factor the quadratic equation:**
We need to factor \( 3x^2 - x - 4 \). To do this, we look for two numbers that multiply to \( 3 \times (-4) = -12 \) and add to \( -1 \). The numbers are \( -4 \) and \( 3 \).
\[ 3x^2 - 4x + 3x - 4 = 0 \]
Now, group the terms:
\[ (3x^2 - 4x) + (3x - 4) = 0 \]
Factor by grouping:
\[ x(3x - 4) + 1(3x - 4) = 0 \]
This gives us:
\[ (3x - 4)(x + 1) = 0 \]

4. **Solve for \( x \):**
Set each factor equal to zero:
\[ 3x - 4 = 0 \quad \text{or} \quad x + 1 = 0 \]
Solving these equations:
- From \( 3x - 4 = 0 \):
\[ 3x = 4 \implies x = \frac{4}{3} \]
- From \( x + 1 = 0 \):
\[ x = -1 \]

5. **Conclusion:**
The values of \( x \) for which the functions are equal are:
\[ x = -1 \quad \text{and} \quad x = \frac{4}{3} \]

### Final Answer:
The domain for which the functions \( f(x) \) and \( g(x) \) are equal is \( x = -1 \) and \( x = \frac{4}{3} \).

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