Question

given $f(x)=\begin{cases}2x - 4, &xleq - 1\ax^{2}-3, &x > - 1end{cases}$, for what value of $a$ is the function continuous at every $x$?

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = - 1\), \(\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)=f(-1)\). First, find \(\lim_{x
ightarrow - 1^{-}}f(x)\).
When \(x
ightarrow - 1^{-}\), \(f(x)=2x - 4\). So \(\lim_{x
ightarrow - 1^{-}}f(x)=2\times(-1)-4=-2 - 4=-6\).

Step2: Find \(\lim_{x

ightarrow - 1^{+}}f(x)\)
When \(x
ightarrow - 1^{+}\), \(f(x)=ax^{2}-3\). So \(\lim_{x
ightarrow - 1^{+}}f(x)=a\times(-1)^{2}-3=a - 3\).

Step3: Set the two - sided limits equal

Since the function is continuous at \(x=-1\), we set \(\lim_{x
ightarrow - 1^{-}}f(x)=\lim_{x
ightarrow - 1^{+}}f(x)\).
\(a - 3=-6\).
Add 3 to both sides of the equation: \(a=-6 + 3=-3\).

Answer:

\(a=-3\)