Determining Values for Function Continuity
The image shows a piecewise function with three different expressions for different intervals of x:
1. \(f(x) = \log_2(3x + a)\) for \(x < 1\)
2. \(f(x) = 3^x\) for \(x = 1\)
3. \(f(x) = \sqrt{x + b}\) for \(x > 1\)
The question asks for the values of \(a\) and \(b\) such that the function \(f(x)\) is continuous at \(x = 1\).
To ensure continuity at \(x = 1\), the following conditions must be met:
1. The limit of \(f(x)\) as \(x\) approaches 1 from the left must equal the value of \(f(1)\).
2. The limit of \(f(x)\) as \(x\) approaches 1 from the right must also equal the value of \(f(1)\).
Let’s find these limits:
1. The limit as \(x\) approaches 1 from the left:
\[
\lim_{{x \to 1^-}} \log_2(3x + a) = \log_2(3 \cdot 1 + a) = \log_2(3 + a)
\]
2. The limit as \(x\) approaches 1 from the right:
\[
\lim_{{x \to 1^+}} \sqrt{x + b} = \sqrt{1 + b}
\]
We also know that \(f(1) = 3^1 = 3\).
Setting these limits equal to \(f(1)\), we get:
\[
\log_2(3 + a) = 3
\]
\[
\sqrt{1 + b} = 3
\]
From the first equation, we can solve for \(a\):
\[
2^3 = 3 + a
\]
\[
8 = 3 + a
\]
\[
a = 5
\]
From the second equation, we can solve for \(b\):
\[
3^2 = 1 + b
\]
\[
9 = 1 + b
\]
\[
b = 8
\]
Therefore, the values of \(a\) and \(b\) are 5 and 8, respectively.
