For what values of x, if any, is the function f(x)= x^2-16/x^2-5x+4 discontinuous

The function \( f(x) = \frac{x^2 - 16}{x^2 - 5x + 4} \) is a rational function, which means it can only be discontinuous where its denominator equals zero. To find the values of \( x \) where the function is discontinuous, we need to solve the equation:

\[
x^2 - 5x + 4 = 0
\]

### Step 1: Solve the quadratic equation \( x^2 - 5x + 4 = 0 \).

We can factor this equation:

\[
x^2 - 5x + 4 = (x - 4)(x - 1) = 0
\]

### Step 2: Find the values of \( x \).

Setting each factor equal to zero, we get:

\[
x - 4 = 0 \quad \text{or} \quad x - 1 = 0
\]

This gives the solutions:

\[
x = 4 \quad \text{or} \quad x = 1
\]

### Step 3: Conclusion

The function is discontinuous at \( x = 4 \) and \( x = 1 \), because the denominator equals zero at these points, making the function undefined there.

Would you like further details on the behavior of the function near these discontinuities?

Here are 5 related questions to explore:
1. How does the function behave as \( x \) approaches 1 or 4 from either side?
2. Can the function be simplified or factored further to understand it better?
3. What are the vertical asymptotes of the function?
4. Are there any removable discontinuities in this function?
5. How can you graph the function and its discontinuities?

**Tip:** Always check the denominator of a rational function to find points of discontinuity.

For what values of x, if any, is the function f(x)= x^2-16/x^2-5x+4 discontinuous?

Explore how to find the values of x that make the rational function f(x) = (x^2 - 16)/(x^2 - 5x + 4) discontinuous. This problem covers algebraic concepts such as solving quadratic equations and understanding discontinuities. Ideal for students in Grades 9-12.

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