Question

for what values of a is the following function continuous at every x?
f(x)=\begin{cases} a^2 x - 2a, & x geq 2 \\ 24, & x < 2 end{cases}
select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. ( a = square ) (simplify your answer. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Recall continuity at a point

For a function to be continuous at \( x = 2 \), the limit as \( x \) approaches 2 from the left (\( \lim_{x \to 2^-} f(x) \)) and the limit as \( x \) approaches 2 from the right (\( \lim_{x \to 2^+} f(x) \)) must equal \( f(2) \). Also, for \( x < 2 \), the function is constant (24), so \( \lim_{x \to 2^-} f(x)=24 \). For \( x \geq 2 \), \( f(x)=a^2x - 2a \), so \( \lim_{x \to 2^+} f(x)=f(2)=a^2(2)-2a = 2a^2 - 2a \).

Step2: Set left and right limits equal

Since the function must be continuous at \( x = 2 \), we set \( \lim_{x \to 2^-} f(x)=\lim_{x \to 2^+} f(x) \), so \( 24 = 2a^2 - 2a \).

Step3: Solve the quadratic equation

Rearrange the equation: \( 2a^2 - 2a - 24 = 0 \). Divide both sides by 2: \( a^2 - a - 12 = 0 \). Factor the quadratic: \( (a - 4)(a + 3)=0 \).

Step4: Find the roots

Set each factor equal to zero: \( a - 4 = 0 \) gives \( a = 4 \), and \( a + 3 = 0 \) gives \( a=-3 \).

Answer:

\( a = -3, 4 \)