Math Problem Statement

Selidiki apakah f kontinu pada titik x=5 untuk fungsi: f(x) = (x^2 - 25)/(x - 5) untuk x ≠ 5, dan f(x) = 6 untuk x = 5.

Solution

To investigate whether the function f(x)f(x) is continuous at x=5x = 5, we need to follow the steps for checking continuity at a point.

Given:

\begin{cases} \frac{x^2 - 25}{x - 5}, & x \neq 5 \\ 6, & x = 5 \end{cases}$$ ### Steps for Continuity: 1. **Check the value of $$f(5)$$:** This is given directly in the definition of the function as $$f(5) = 6$$. 2. **Evaluate the limit of $$f(x)$$ as $$x$$ approaches 5:** We need to compute: $$\lim_{x \to 5} f(x) = \lim_{x \to 5} \frac{x^2 - 25}{x - 5}$$ Notice that $$x^2 - 25$$ can be factored: $$\frac{x^2 - 25}{x - 5} = \frac{(x - 5)(x + 5)}{x - 5}$$ For $$x \neq 5$$, the $$(x - 5)$$ terms cancel out, leaving: $$\frac{(x - 5)(x + 5)}{x - 5} = x + 5$$ So, $$\lim_{x \to 5} f(x) = \lim_{x \to 5} (x + 5) = 10$$. 3. **Compare $$f(5)$$ with $$\lim_{x \to 5} f(x)$$:** We find that: $$\lim_{x \to 5} f(x) = 10 \quad \text{and} \quad f(5) = 6$$ Since these values are not equal, the function $$f(x)$$ is **not continuous** at $$x = 5$$. ### Conclusion: The function $$f(x)$$ is **discontinuous** at $$x = 5$$ because $$\lim_{x \to 5} f(x) \neq f(5)$$. --- Would you like a more detailed explanation or have any questions? Here are some related questions for further understanding: 1. What is the general condition for a function to be continuous at a point? 2. How do we handle discontinuities of different types (e.g., jump discontinuities)? 3. How can factoring help in finding limits like the one we solved? 4. What does it mean for a function to be "piecewise-defined"? 5. What are the potential consequences of discontinuities in real-world applications? **Tip:** Always try to simplify expressions before evaluating limits, especially when dealing with indeterminate forms like $$\frac{0}{0}$$.

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Math Problem Analysis

Mathematical Concepts

Continuity
Limits
Piecewise Function
Factoring

Formulas

Limit of a function as x approaches a value
Factoring: x^2 - 25 = (x - 5)(x + 5)

Theorems

Continuity Theorem: A function is continuous at x = a if \( \lim_{x \to a} f(x) = f(a) \)

Suitable Grade Level

Grades 10-12