Math Problem Statement
Determine a value of c such that lim(x→2) g(x) exists, where g(x) is a piecewise function defined as: g(x) = {x^2 + cx, for x < 2; c, for x = 2; 3x + 4c, for x > 2}
Solution
We are asked to determine the value of such that exists, where is defined piecewise as follows:
x^2 + cx, & x < 2 \\ c, & x = 2 \\ 3x + 4c, & x > 2 \end{cases}$$ For the limit to exist at $$x = 2$$, the left-hand limit ($$\lim_{x \to 2^-} g(x)$$) and the right-hand limit ($$\lim_{x \to 2^+} g(x)$$) must be equal to the function value at $$x = 2$$ (which is $$g(2) = c$$). ### 1. Left-hand limit ($$x \to 2^-$$): For $$x < 2$$, the function is $$g(x) = x^2 + cx$$. Taking the limit as $$x \to 2$$: $$\lim_{x \to 2^-} g(x) = 2^2 + c \cdot 2 = 4 + 2c$$ ### 2. Right-hand limit ($$x \to 2^+$$): For $$x > 2$$, the function is $$g(x) = 3x + 4c$$. Taking the limit as $$x \to 2$$: $$\lim_{x \to 2^+} g(x) = 3(2) + 4c = 6 + 4c$$ ### 3. Equating limits: For the limit to exist, we require that the left-hand limit, right-hand limit, and the value of the function at $$x = 2$$ all be equal: $$4 + 2c = c$$ $$6 + 4c = c$$ We solve each equation for $$c$$. #### Solving $$4 + 2c = c$$: $$4 + 2c = c \implies c = -4$$ #### Solving $$6 + 4c = c$$: $$6 + 4c = c \implies 6 = -3c \implies c = -2$$ ### Conclusion: There is no single value of $$c$$ that satisfies both conditions simultaneously. Therefore, no value of $$c$$ ensures that the limit exists. Would you like further details on this process? Here are some related questions to explore: 1. What happens when we evaluate a limit at a discontinuous point? 2. Can limits exist for piecewise functions with different conditions on either side? 3. How can we use continuity to solve similar problems? 4. What is the difference between left-hand and right-hand limits in general? 5. How does a piecewise function’s structure affect its differentiability? **Tip:** When dealing with limits of piecewise functions, always check both the left-hand and right-hand limits carefully to ensure they match.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise functions
Limits
Continuity
Formulas
lim(x→2-) g(x) = 2^2 + 2c = 4 + 2c
lim(x→2+) g(x) = 3(2) + 4c = 6 + 4c
Theorems
Continuity theorem
Limit theorem for piecewise functions
Suitable Grade Level
Grades 11-12