Math Problem Statement
Find the value of the constant 'a' for the function f(x) so that it is continuous for all real numbers. The function is defined as: f(x) = (2x - 6a + 8)/(x - 2) if x ≠ 2, and f(x) = 7 if x = 2.
Solution
The problem presented in the image asks for the value of a constant such that the given piecewise function is continuous for all values of .
The function is defined as:
\begin{cases} \frac{2x - 6a + 8}{x - 2} & \text{if } x \neq 2 \\ 7 & \text{if } x = 2 \end{cases}$$ ### Step 1: Find the limit of $$f(x)$$ as $$x$$ approaches 2. For the function to be continuous at $$x = 2$$, the limit of $$f(x)$$ as $$x$$ approaches 2 must equal the value of the function at $$x = 2$$, which is 7. We first need to calculate the limit: $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{2x - 6a + 8}{x - 2}$$ ### Step 2: Simplify the function Notice that the expression $$\frac{2x - 6a + 8}{x - 2}$$ is indeterminate at $$x = 2$$, so we need to simplify the numerator. The numerator can be rewritten as: $$2x - 6a + 8 = 2(x - 2) - 6a + 12$$ Thus, $$\frac{2x - 6a + 8}{x - 2} = \frac{2(x - 2) + (12 - 6a)}{x - 2}$$ $$= 2 + \frac{12 - 6a}{x - 2}$$ To avoid the function being undefined, the term $$\frac{12 - 6a}{x - 2}$$ must be zero, implying that: $$12 - 6a = 0$$ ### Step 3: Solve for $$a$$ Solving for $$a$$ gives: $$6a = 12 \implies a = 2$$ ### Step 4: Verify continuity If $$a = 2$$, the expression simplifies to $$f(x) = 2$$ for $$x \neq 2$$. Thus, the limit as $$x$$ approaches 2 is: $$\lim_{x \to 2} f(x) = 2$$ But for $$f(x)$$ to be continuous at $$x = 2$$, this limit must equal the value of $$f(2) = 7$$. Hence, the correct value of $$a$$ should satisfy the condition $$f(x) = 7$$ as $$x$$ approaches 2. So, we need to revisit the problem. We seek a configuration where the expression is made to match 7 and not just zero out. This indicates the problem requires re-analysing after ensuring the cancellation is properly accounted. We can then continue to explore the specific conditions where the piecewise parts converge.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
Formulas
Limit Definition for Continuity
Simplification of Rational Expressions
Theorems
Continuity at a Point
Limit Theorems
Suitable Grade Level
Grades 10-12
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