Question

find the value of the constant m that makes the following function continuous on (-∞,∞).
f(x)=\begin{cases}mx - 10&\text{if }x < - 8\\x^{2}+6x - 2&\text{if }xgeq - 8end{cases}
now draw a graph of f.

Explanation:

Step1: Recall continuity condition

For a function to be continuous at $x = - 8$, $\lim_{x
ightarrow - 8^{-}}f(x)=\lim_{x
ightarrow - 8^{+}}f(x)$.

Step2: Calculate left - hand limit

$\lim_{x
ightarrow - 8^{-}}f(x)=\lim_{x
ightarrow - 8^{-}}(mx - 10)=m(-8)-10=-8m - 10$.

Step3: Calculate right - hand limit

$\lim_{x
ightarrow - 8^{+}}f(x)=\lim_{x
ightarrow - 8^{+}}(x^{2}+6x - 2)=(-8)^{2}+6(-8)-2=64-48 - 2=14$.

Step4: Set left - hand equal to right - hand limit

Set $-8m - 10 = 14$. Add 10 to both sides: $-8m=14 + 10=24$. Then divide by - 8, so $m=-3$.

Answer:

$-3$