Question - Determining Values for Function Continuity
Solution:
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.
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