Finding Values for Continuity in a Piecewise Function
The image shows a mathematics question regarding a piecewise-defined function. The function \( f(x) \) is defined differently for different intervals of \( x \):
\[ f(x) = \begin{cases}
\log_3(3x + a) & , x < 1 \\
3x & , x = 1 \\
\sqrt{x + b} & , x > 1
\end{cases} \]
The question states that the function is continuous at \( x = 1 \) and asks for the values of \( a \) and \( b \).
Continuity at \( x = 1 \) means that the left-hand limit as \( x \) approaches 1, the right-hand limit as \( x \) approaches 1, and the function's value at \( x = 1 \) must all be equal. Let's calculate these:
1. The left-hand limit as \( x \) approaches 1 is the limit of \( \log_3(3x + a) \) as \( x \) approaches 1 from the left:
\[ \lim_{{x \to 1^-}} \log_3(3x + a) = \log_3(3 \cdot 1 + a) = \log_3(3 + a) \]
2. The function's value at \( x = 1 \) is given directly as \( f(1) = 3 \cdot 1 = 3 \).
3. The right-hand limit as \( x \) approaches 1 is the limit of \( \sqrt{x + b} \) as \( x \) approaches 1 from the right:
\[ \lim_{{x \to 1^+}} \sqrt{x + b} = \sqrt{1 + b} \]
For continuity at \( x = 1 \), we must have \( \log_3(3 + a) = 3 = \sqrt{1 + b} \). From here we can find the values of \( a \) and \( b \):
\(\log_3(3 + a) = 3\) implies \(3^3 = 3 + a\), hence \( a = 27 - 3 = 24 \).
\(\sqrt{1 + b} = 3\) implies \(1 + b = 3^2\), hence \( b = 9 - 1 = 8 \).
Thus, \( a = 24 \) and \( b = 8 \).
