Questions: Question 9 Let f(x) = mx-12 if x < -6 x^2 + 5x - 6 if x ≥ -6 If f(x) is a function which is continuous everywhere, then we must have m =

Transcript text: Question 9 Let $f(x)=\left\{\begin{array}{ll}m x-12 & \text { if } x<-6 \\ x^{2}+5 x-6 & \text { if } x \geq-6\end{array}\right.$ If $f(x)$ is a function which is continuous everywhere, then we must have \[ m= \] Question Help: Video

Solution

Solution Steps

To ensure the function $ f(x) $ is continuous everywhere, the left-hand limit as $ x $ approaches $-6$ from the left must equal the right-hand limit as $ x $ approaches $-6$ from the right. This means we need to set the two expressions for $ f(x) $ equal to each other at $ x = -6 $ and solve for $ m $.

Step 1: Understand the Problem

We are given a piecewise function $ f(x) $ and need to determine the value of $ m $ such that $ f(x) $ is continuous everywhere. The function is defined as: \[ f(x) = \begin{cases} m x - 12 & \text{if } x < -6 \\ x^2 + 5x - 6 & \text{if } x \geq -6 \end{cases} \] To ensure continuity at $ x = -6 $, the left-hand limit as $ x $ approaches $-6$ must equal the right-hand limit and the function value at $ x = -6 $.

Step 2: Calculate the Left-Hand Limit

The left-hand limit of $ f(x) $ as $ x $ approaches $-6$ is given by the expression for $ x < -6 $: \[ \lim_{x \to -6^-} f(x) = m(-6) - 12 = -6m - 12 \]

Step 3: Calculate the Right-Hand Limit and Function Value

The right-hand limit of $ f(x) $ as $ x $ approaches $-6$ is given by the expression for $ x \geq -6 $: \[ \lim_{x \to -6^+} f(x) = (-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0 \] The function value at $ x = -6 $ is: \[ f(-6) = (-6)^2 + 5(-6) - 6 = 36 - 30 - 6 = 0 \]

Step 4: Set the Limits Equal for Continuity

For $ f(x) $ to be continuous at $ x = -6 $, the left-hand limit must equal the right-hand limit and the function value: \[ -6m - 12 = 0 \]

Step 5: Solve for $ m $

Solve the equation for $ m $: \[ -6m - 12 = 0 \\ -6m = 12 \\ m = -2 \]