6. Find \( (f \circ g)(x) \) when \( f(x)=x^{2}-1 \) and \( g(x)=\sqrt{1-3 x} \) and state the domain of the composition

To find the composition \( (f \circ g)(x) \), we need to substitute \( g(x) \) into \( f(x) \). Given: - \( f(x) = x^2 - 1 \) - \( g(x) = \sqrt{1 - 3x} \) The composition \( (f \circ g)(x) \) is defined as: \[ (f \circ g)(x) = f(g(x)) = f(\sqrt{1 - 3x}) \] Now, we substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f(\sqrt{1 - 3x}) = (\sqrt{1 - 3x})^2 - 1 \] Calculating this gives: \[ (\sqrt{1 - 3x})^2 = 1 - 3x \] Thus, \[ f(g(x)) = 1 - 3x - 1 = -3x \] So, we have: \[ (f \circ g)(x) = -3x \] Next, we need to determine the domain of the composition \( (f \circ g)(x) \). The domain of \( (f \circ g)(x) \) is determined by the domain of \( g(x) \) and the values that \( g(x) \) can take that are valid inputs for \( f(x) \). 1. **Domain of \( g(x) \)**: The function \( g(x) = \sqrt{1 - 3x} \) is defined when the expression under the square root is non-negative: \[ 1 - 3x \geq 0 \] Solving this inequality: \[ 1 \geq 3x \quad \Rightarrow \quad x \leq \frac{1}{3} \] Therefore, the domain of \( g(x) \) is \( (-\infty, \frac{1}{3}] \). 2. **Range of \( g(x) \)**: The output of \( g(x) \) must also be valid for \( f(x) \). Since \( g(x) \) outputs non-negative values (as it is a square root), we have: \[ g(x) \geq 0 \quad \Rightarrow \quad \sqrt{1 - 3x} \geq 0 \] This is satisfied for all \( x \) in the domain of \( g(x) \). 3. **Domain of \( f(x) \)**: The function \( f(x) = x^2 - 1 \) is defined for all real numbers, so there are no additional restrictions from \( f(x) \). Combining these, the domain of \( (f \circ g)(x) \) is simply the domain of \( g(x) \): \[ \text{Domain of } (f \circ g)(x) = (-\infty, \frac{1}{3}] \] In conclusion, we have: \[ (f \circ g)(x) = -3x \] and the domain of the composition is: \[ \text{Domain: } (-\infty, \frac{1}{3}] \]