\( f(x)=\left\{\begin{array}{ll}x^{2}-1 & , x \leq-2 \\ m(x-2)-1 & , x>-2\end{array}\right. \) is continuous everywhere, then we must have \( m=\square \) Consequently, \( f(0)=\square \)

To ensure that the function \( f(x) \) is continuous everywhere, we need to check the continuity at the point where the definition of the function changes, which is at \( x = -2 \). The function is defined as follows: \[ f(x) = \begin{cases} x^2 - 1 & \text{if } x \leq -2 \\ m(x - 2) - 1 & \text{if } x > -2 \end{cases} \] 1. **Calculate \( f(-2) \)** using the first piece of the function: \[ f(-2) = (-2)^2 - 1 = 4 - 1 = 3 \] 2. **Calculate the limit of \( f(x) \) as \( x \) approaches -2 from the right** (using the second piece of the function): \[ \lim_{x \to -2^+} f(x) = m(-2 - 2) - 1 = m(-4) - 1 = -4m - 1 \] 3. **Set the two expressions equal to ensure continuity at \( x = -2 \)**: \[ f(-2) = \lim_{x \to -2^+} f(x) \] This gives us the equation: \[ 3 = -4m - 1 \] 4. **Solve for \( m \)**: \[ 3 + 1 = -4m \\ 4 = -4m \\ m = -1 \] Thus, we have \( m = -1 \). 5. **Now, we can find \( f(0) \)**. Since \( 0 > -2 \), we use the second piece of the function: \[ f(0) = m(0 - 2) - 1 = -1(0 - 2) - 1 = -1(-2) - 1 = 2 - 1 = 1 \] In conclusion, we have: \[ m = -1 \quad \text{and} \quad f(0) = 1 \] So, the final answers are: \[ m = -1 \quad \text{and} \quad f(0) = 1 \]