Question

for what values of a is the following function continuous at every x?
f(x)=\begin{cases}a^{2}x - 4a, & xgeq2\\16, & x < 2end{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

A function is continuous at $x = c$ if $\lim_{x
ightarrow c^{-}}f(x)=\lim_{x
ightarrow c^{+}}f(x)=f(c)$. Here $c = 2$, so $\lim_{x
ightarrow 2^{-}}f(x)=\lim_{x
ightarrow 2^{+}}f(x)$.

Step2: Calculate left - hand limit

For $x<2$, $f(x)=16$. So $\lim_{x
ightarrow 2^{-}}f(x)=16$.

Step3: Calculate right - hand limit

For $x\geq2$, $f(x)=a^{2}x - 4a$. Then $\lim_{x
ightarrow 2^{+}}f(x)=a^{2}(2)-4a=2a^{2}-4a$.

Step4: Set left - hand and right - hand limits equal

Set $2a^{2}-4a = 16$. Rearrange to get $2a^{2}-4a - 16=0$. Divide by 2: $a^{2}-2a - 8=0$.

Step5: Solve the quadratic equation

Factor the quadratic equation $a^{2}-2a - 8=(a - 4)(a+2)=0$. So $a - 4=0$ or $a + 2=0$, which gives $a = 4$ or $a=-2$.

Answer:

A. $a = 4,-2$