Question

for what value of a is the following function continuous at every x?
$f(x) = \

$$\begin{cases} x^2 - 4, & x < 3 \\\\ 2ax, & x \\geq 3 \\end{cases}$$

$
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
\\(\bigcirc\\) a. \\(a = \square\\) (simplify your answer. use a comma to separate answers as needed)
\\(\bigcirc\\) b. there is no solution.

Explanation:

Step1: Find left - hand limit at x = 3

For \(x\lt3\), \(f(x)=x^{2}-4\). The left - hand limit as \(x\) approaches 3 is \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}(x^{2}-4)\). Substitute \(x = 3\) into \(x^{2}-4\), we get \(3^{2}-4=9 - 4=5\).

Step2: Find right - hand limit at x = 3

For \(x\geq3\), \(f(x)=2ax\). The right - hand limit as \(x\) approaches 3 is \(\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}(2ax)\). Substitute \(x = 3\) into \(2ax\), we get \(2a\times3 = 6a\).

Step3: Set left - hand limit equal to right - hand limit for continuity

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\). So we set \(5 = 6a\).

Step4: Solve for a

To solve for \(a\), we divide both sides of the equation \(5 = 6a\) by 6. We get \(a=\frac{5}{6}\).

Answer:

\(a=\frac{5}{6}\)