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

for what value of a is the following function continuous at every x?\n f(x)=\begin{cases}x^{2}-4, & x < 3\\2ax, & xgeq3end{cases}\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. a = (simplify your answer. use a comma to separate answers as needed.)\nb. there is no solution.

for what value of a is the following function continuous at every x?\n f(x)=\begin{cases}x^{2}-4, & x < 3\\2ax, & xgeq3end{cases}\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. a = (simplify your answer. use a comma to separate answers as needed.)\nb. there is no solution.

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at (x = 3), (\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)=f(3)). First, find (\lim_{x\rightarrow3^{-}}f(x)). When (x\rightarrow3^{-}), (f(x)=x^{2}-4). So (\lim_{x\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{-}}(x^{2}-4)). Substitute (x = 3) into (x^{2}-4): (3^{2}-4=9 - 4=5).

Step2: Find right - hand limit

When (x\rightarrow3^{+}), (f(x)=2ax). So (\lim_{x\rightarrow3^{+}}f(x)=\lim_{x\rightarrow3^{+}}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\rightarrow3^{-}}f(x)=\lim_{x\rightarrow3^{+}}f(x)). So (6a=5). Solve for (a): (a=\frac{5}{6}).

Answer:

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