Math Problem Statement
Given the piecewise function: f(x) = { sqrt(kx), 0 ≤ x ≤ 3; x + 1, 3 < x ≤ 10 }, find the value of k that ensures continuity at x = 3.
Solution
The piecewise function provided is:
\begin{cases} \sqrt{kx}, & 0 \leq x \leq 3 \\ x + 1, & 3 < x \leq 10 \end{cases}$$ To ensure continuity at $$x = 3$$, the two parts of the function must have the same value at that point. Therefore, we need to set: $$\sqrt{k \cdot 3} = 3 + 1$$ Now, let's solve for $$k$$. First, simplify both sides of the equation: $$\sqrt{3k} = 4$$ Square both sides: $$3k = 16$$ Now solve for $$k$$: $$k = \frac{16}{3}$$ So, for the function to be continuous at $$x = 3$$, $$k = \frac{16}{3}$$. Would you like me to go into more details, or do you have any further questions? Here are some related questions: 1. How do we check if a function is differentiable at a point where two pieces meet? 2. Can piecewise functions always be made continuous by adjusting parameters? 3. What does it mean for a function to be continuous at a point? 4. Can you find the derivative of this piecewise function? 5. How does the square root affect the domain of the function? **Tip:** When solving for continuity in piecewise functions, always equate the boundary values of the two expressions at the connecting point!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Formulas
sqrt(kx) for 0 ≤ x ≤ 3
x + 1 for 3 < x ≤ 10
Continuity condition: sqrt(k*3) = 3 + 1
Theorems
Continuity Theorem
Suitable Grade Level
Grades 9-12