Question

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

$$\begin{cases}x^2 - 7, & x<3 \\\\ 2ax, & x\\geq 3\\end{cases}$$

$
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 the left - hand limit at x = 3

For \(x\lt3\), the function is \(f(x)=x^{2}-7\). The left - hand limit as \(x\) approaches 3 is \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}(x^{2}-7)\).
Substitute \(x = 3\) into \(x^{2}-7\), we get \(3^{2}-7=9 - 7=2\).

Step2: Find the right - hand limit (and the function value) at x = 3

For \(x\geq3\), the function is \(f(x)=2ax\). 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 \(2a\times3 = 6a\). Also, \(f(3)=2a\times3=6a\).

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

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).
We know that \(\lim_{x
ightarrow3^{-}}f(x) = 2\) and \(\lim_{x
ightarrow3^{+}}f(x)=6a\). So we set up the equation \(6a=2\).
Solve for \(a\): Divide both sides of the equation \(6a = 2\) by 6, we get \(a=\frac{2}{6}=\frac{1}{3}\).

Answer:

A. \(a=\frac{1}{3}\)