If f(x)=x(x-2)/x-1 then find f(0)+f(-1)+f(-4)+f(2)-5/2

We are given the function:

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

and are tasked to find:

\[
f(0) + f(-1) + f(-4) + f(2) - \frac{5}{2}
\]

Let's evaluate the function at each value of \( x \) separately:

### 1. \( f(0) \)
\[
f(0) = \frac{0(0-2)}{0-1} = \frac{0}{-1} = 0
\]

### 2. \( f(-1) \)
\[
f(-1) = \frac{-1(-1-2)}{-1-1} = \frac{-1(-3)}{-2} = \frac{3}{-2} = -\frac{3}{2}
\]

### 3. \( f(-4) \)
\[
f(-4) = \frac{-4(-4-2)}{-4-1} = \frac{-4(-6)}{-5} = \frac{24}{-5} = -\frac{24}{5}
\]

### 4. \( f(2) \)
Here, \( f(2) \) is undefined because the denominator \( x - 1 \) becomes zero when \( x = 2 \). Therefore, \( f(2) \) is not defined.

### Putting it all together:
Since \( f(2) \) is undefined, the entire expression becomes undefined.

Thus, the final result is **undefined** due to \( f(2) \).

Would you like more details or have any questions about this?

Here are 5 related questions:
1. What is the behavior of the function \( f(x) \) as \( x \to 1 \)?
2. How would you handle limits in this case?
3. Can the discontinuity at \( x = 1 \) be removed by redefining \( f(x) \)?
4. How would you compute the limit of \( f(x) \) as \( x \to 2 \)?
5. What is the domain of the function \( f(x) \)?

**Tip:** Always check the domain of a function before evaluating it at specific points to avoid undefined values.

If f(x) = x(x-2)/(x-1), then find f(0) + f(-1) + f(-4) + f(2) - 5/2.

Learn how to evaluate a rational function f(x) = x(x-2)/(x-1) at different points, including handling undefined values and discontinuities. This problem explores key algebraic concepts suitable for students in Grades 9-11.

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