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The problem asks why the function \( f(x) \) is not continuous at \( x = 1 \). To determine this, let's analyze the given function:

\[
f(x) = \frac{x^2 + x - 2}{x^2 - 5x + 4}
\]

### Step 1: Factorize the numerator and the denominator.
- **Numerator**: \( x^2 + x - 2 \) can be factored as \( (x - 1)(x + 2) \).
- **Denominator**: \( x^2 - 5x + 4 \) can be factored as \( (x - 1)(x - 4) \).

So the function can be rewritten as:

\[
f(x) = \frac{(x - 1)(x + 2)}{(x - 1)(x - 4)}
\]

### Step 2: Simplify the function.
- The factor \( (x - 1) \) appears in both the numerator and denominator, so it can be canceled out for all \( x \neq 1 \).
  
Thus, for \( x \neq 1 \), the function simplifies to:

\[
f(x) = \frac{x + 2}{x - 4}
\]

### Step 3: Evaluate the conditions for continuity at \( x = 1 \).

For a function to be continuous at \( x = 1 \), the following three conditions must be satisfied:
1. \( \lim_{x \to 1} f(x) \) exists.
2. \( f(1) \) is defined.
3. \( \lim_{x \to 1} f(x) = f(1) \).

- **Limit**: To find \( \lim_{x \to 1} f(x) \), we use the simplified function \( f(x) = \frac{x + 2}{x - 4} \). As \( x \) approaches 1:
  \[
  \lim_{x \to 1} f(x) = \frac{1 + 2}{1 - 4} = \frac{3}{-3} = -1
  \]
  
- **Function value**: To find \( f(1) \), substitute \( x = 1 \) into the original function:
  \[
  f(1) = \frac{1^2 + 1 - 2}{1^2 - 5(1) + 4} = \frac{0}{0}
  \]
  The result \( \frac{0}{0} \) is undefined, indicating that \( f(1) \) does not exist.

Since \( f(1) \) does not exist, the function is not continuous at \( x = 1 \).

### Answer:
The correct choice is **B**: \( \lim_{x \to 1} f(x) \) exists, but \( f(1) \) does not exist.

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Would you like more details or have any questions?

### 5 Related Questions:
1. How can you determine if a function is continuous at a different point, say \( x = 2 \)?
2. What is the significance of removing a removable discontinuity?
3. How would the analysis change if \( x = 4 \) was the point of interest?
4. Can you always cancel out factors like \( (x - 1) \) in any rational function?
5. What are other common types of discontinuities besides removable discontinuities?

### Tip:
Always check if a function value is defined at the point of interest when testing for continuity, especially if there's an indeterminate form like \( \frac{0}{0} \).

Let f be the function defined above. Which of the following conditions explains why f is not continuous at x = 1?

Explore why the function f(x) = (x^2 + x - 2)/(x^2 - 5x + 4) is not continuous at x = 1. This problem highlights key mathematical concepts such as continuity, limits, and factorization. Ideal for students in Grades 11-12.

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