Math Problem Statement
Let g(x) be a piecewise function defined as follows: g(x) = (x-3)/(sqrt(x+13)-4) for x ≥ -13, x ≠ 3, and g(x) = k for x = 3. The function g is continuous for all x > -13. What is the value of k?
Solution
To solve this problem, we need to determine the value of such that the function is continuous for all .
Step 1: Understand the given function
The function is defined as:
\frac{x - 3}{\sqrt{x + 13} - 4} & \text{for } x \geq -13, x \neq 3 \\ k & \text{for } x = 3 \end{cases}$$ The function is continuous for all $$x > -13$$. For $$g(x)$$ to be continuous at $$x = 3$$, the value of $$g(x)$$ as $$x \to 3$$ (from the piecewise function definition) must equal $$g(3) = k$$. ### Step 2: Find the limit as $$x \to 3$$ To make $$g(x)$$ continuous at $$x = 3$$, we must find the limit of $$\frac{x - 3}{\sqrt{x + 13} - 4}$$ as $$x \to 3$$. Substituting $$x = 3$$ into the function directly gives a $$\frac{0}{0}$$ indeterminate form, so we apply algebraic simplification. #### Simplifying the expression: $$\lim_{x \to 3} \frac{x - 3}{\sqrt{x + 13} - 4}$$ Multiply the numerator and denominator by the conjugate of the denominator: $$\lim_{x \to 3} \frac{(x - 3)(\sqrt{x + 13} + 4)}{(\sqrt{x + 13} - 4)(\sqrt{x + 13} + 4)}$$ This simplifies to: $$\lim_{x \to 3} \frac{(x - 3)(\sqrt{x + 13} + 4)}{(x + 13) - 16} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x + 13} + 4)}{x - 3}$$ Cancel out the $$(x - 3)$$ terms: $$\lim_{x \to 3} \sqrt{x + 13} + 4$$ Now substitute $$x = 3$$ into the simplified expression: $$\sqrt{3 + 13} + 4 = \sqrt{16} + 4 = 4 + 4 = 8$$ ### Step 3: Set the limit equal to $$k$$ For $$g(x)$$ to be continuous at $$x = 3$$, the value of $$k$$ must equal the limit as $$x \to 3$$. Therefore, $$k = 8$$. ### Final Answer: The value of $$k$$ is $$\boxed{8}$$. --- Would you like further details or explanations on any step? Here are some related questions to explore: 1. What does it mean for a function to be continuous at a point? 2. How do you use limits to test for continuity? 3. Why did we multiply by the conjugate in this problem? 4. Can you always simplify indeterminate forms like $$\frac{0}{0}$$ using algebraic techniques? 5. How does the value of $$k$$ affect the overall behavior of the function? **Tip**: When you encounter a $$\frac{0}{0}$$ indeterminate form, applying algebraic tricks like multiplying by the conjugate often helps to simplify the limit.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Limits
Continuity
Algebraic Simplification
Formulas
Limit as x approaches a value
Multiplying by the conjugate
Continuity condition: limit of f(x) = f(c)
Theorems
Limit Theorem
Squeeze Theorem
Properties of Continuous Functions
Suitable Grade Level
Grades 10-12