Question

for what value of a is the following function continuous at every x?
f(x)=\begin{cases}x^{2}-4, & x < 3\\2ax, & xgeq3end{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

For a function to be continuous at \(x = 3\), \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)=f(3)\). First, find \(\lim_{x
ightarrow3^{-}}f(x)\).
When \(x
ightarrow3^{-}\), \(f(x)=x^{2}-4\). So \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}(x^{2}-4)\).
Substitute \(x = 3\) into \(x^{2}-4\): \(3^{2}-4=9 - 4=5\).

Step2: Find right - hand limit

When \(x
ightarrow3^{+}\), \(f(x)=2ax\). So \(\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}2ax = 2a\times3=6a\).

Step3: Set left - hand and right - hand limits equal

Since the function is continuous at \(x = 3\), we set \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).
So \(6a=5\).
Solve for \(a\): \(a=\frac{5}{6}\).

Answer:

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