for what values of a is the following function continuous at every x?\n f(x)=\begin{cases}a^{2}x - 3a…

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

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

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at (x = 3), (\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)). Here, (\lim_{x\rightarrow3^{-}}f(x) = 18) and (\lim_{x\rightarrow3^{+}}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)