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

Explanation:

Step1: Recall continuity condition

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

Step2: Find (\lim_{x\rightarrow - 1^{+}}f(x))

When (x\rightarrow - 1^{+}), (f(x)=ax^{2}-3). So (\lim_{x\rightarrow - 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\rightarrow - 1^{-}}f(x)=\lim_{x\rightarrow - 1^{+}}f(x)). (a - 3=-6). Add 3 to both sides of the equation: (a=-6 + 3=-3).

Answer:

(a=-3)