Let \( f(x)=\left\{\begin{array}{ll}m x-15 & \text { if } \quad x<-10 \\ x^{2}+8 x-5 & \text { if } \quad x \geq-10\end{array}\right. \) If \( f(x) \) is a function which is continuous everywhere, then we must have \( m=\square \) Question Help: \( \square \) Video \( \square \) Message instructor Submit Question Jump to Answer

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 \( x = -10 \). The function is defined as follows: \[ f(x) = \begin{cases} m x - 15 & \text{if } x < -10 \\ x^2 + 8x - 5 & \text{if } x \geq -10 \end{cases} \] To check for continuity at \( x = -10 \), we need to ensure that: \[ \lim_{x \to -10^-} f(x) = \lim_{x \to -10^+} f(x) = f(-10) \] 1. **Calculate \( f(-10) \)** using the second piece of the function (since \( -10 \geq -10 \)): \[ f(-10) = (-10)^2 + 8(-10) - 5 = 100 - 80 - 5 = 15 \] 2. **Calculate \( \lim_{x \to -10^-} f(x) \)** using the first piece of the function: \[ \lim_{x \to -10^-} f(x) = m(-10) - 15 = -10m - 15 \] 3. **Calculate \( \lim_{x \to -10^+} f(x) \)** using the second piece of the function: \[ \lim_{x \to -10^+} f(x) = f(-10) = 15 \] For continuity at \( x = -10 \), we set the left-hand limit equal to the value of the function at that point: \[ -10m - 15 = 15 \] Now, solve for \( m \): \[ -10m - 15 = 15 \\ -10m = 15 + 15 \\ -10m = 30 \\ m = -3 \] Thus, the value of \( m \) that makes the function continuous everywhere is: \[ \boxed{-3} \]