Question

question 7
let $f(x)=\

$$\begin{cases}mx - 6&\\text{if }x < -3\\\\x^2 + 8x - 3&\\text{if }x \\geq -3\\end{cases}$$

$
if $f(x)$ is a function which is continuous everywhere, then we must have
$m = \square$
question help: video

Explanation:

Step1: Find right-hand limit at $x=-3$

For $x\geq-3$, use $f(x)=x^2+8x-3$.
$\lim_{x\to -3^+} f(x) = (-3)^2 + 8(-3) - 3$
$= 9 - 24 - 3 = -18$

Step2: Find left-hand limit at $x=-3$

For $x<-3$, use $f(x)=mx-6$.
$\lim_{x\to -3^-} f(x) = m(-3) - 6 = -3m - 6$

Step3: Set limits equal for continuity

For continuity at $x=-3$, left and right limits must be equal.
$-3m - 6 = -18$

Step4: Solve for $m$

Rearrange and solve the equation:
$-3m = -18 + 6$
$-3m = -12$
$m = \frac{-12}{-3} = 4$

Answer:

$4$