Question - Finding Values for Continuity in a Piecewise Function
Solution:
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 $$ b = 8 $$.
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