Where is the function \( f(x)=\frac{x^{2}-16}{x^{2}-6 x+8} \) continuous?

1. Factorize the numerator and the denominator: - Numerator: \[ x^2 - 16 = (x - 4)(x + 4) \] - Denominator: \[ x^2 - 6x + 8 = (x - 2)(x - 4) \] 2. Simplify the function by canceling the common factor \((x - 4)\). Note that this cancellation is valid provided that \((x - 4) \neq 0\), i.e., \(x \neq 4\): \[ f(x) = \frac{(x - 4)(x + 4)}{(x - 2)(x - 4)} = \frac{x + 4}{x - 2} \quad \text{for} \quad x \neq 4. \] 3. Determine the points where the function is not originally defined by setting the denominator equal to zero: \[ (x - 2)(x - 4) = 0 \implies x = 2 \quad \text{or} \quad x = 4. \] Therefore, the function is not defined at \(x = 2\) and \(x = 4\). 4. The function \( f(x) \) is continuous wherever it is defined. Since discontinuities occur at \(x = 2\) and \(x = 4\), the function is continuous for all \(x \in \mathbb{R}\) except at those points. Thus, the function \( f(x) = \frac{x^2 - 16}{x^2 - 6x + 8} \) is continuous on: \[ \{ x \in \mathbb{R} : x \neq 2 \text{ and } x \neq 4 \}. \]