Question

for what values of a is the following function continuous at every x?
f(x)=\begin{cases}a^{2}x - 3a, &xgeq3\\18, &x < 3end{cases}
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. a = (simplify your answer. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\). Here, \(\lim_{x
ightarrow3^{-}}f(x) = 18\) and \(\lim_{x
ightarrow3^{+}}f(x)=a^{2}\times3 - 3a\).

Step2: Set up the equation

Set \(3a^{2}-3a=18\). Rearrange it to the standard - quadratic form \(3a^{2}-3a - 18 = 0\). Divide through by 3 to get \(a^{2}-a - 6=0\).

Step3: Solve the quadratic equation

Factor the quadratic equation \(a^{2}-a - 6=(a - 3)(a + 2)=0\). Using the zero - product property, if \((a - 3)(a + 2)=0\), then \(a-3 = 0\) or \(a + 2=0\).

Answer:

\(a=-2,3\)