Question

let $f(x)=\begin{cases}mx - 16&\text{if }x < - 10\\x^{2}+7x - 6&\text{if }xgeq - 10end{cases}$. if $f(x)$ is a function which is continuous everywhere, then $m=$

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = - 10$, $\lim_{x
ightarrow - 10^{-}}f(x)=\lim_{x
ightarrow - 10^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow - 10^{-}}f(x)=m(-10)-16=-10m - 16$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow - 10^{+}}f(x)=(-10)^{2}+7(-10)-6=100 - 70-6 = 24$.

Step4: Set left - hand and right - hand limits equal

$-10m - 16=24$.

Step5: Solve for $m$

Add 16 to both sides: $-10m=24 + 16=40$. Then divide by - 10, $m=\frac{40}{-10}=-4$.

Answer:

$-4$