Find all points of discontinuity for the following piecewise function:
f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪−2x, if x<−2x2, if −2≤x<03x, if 0<x<3xx−4, if 3≤x≤6xx−5, if 6<x≤7x2, if 7<x

To find all points of discontinuity for the given piecewise function:

\[
f(x) = 
\begin{cases} 
-2x & \text{if } x < -2 \\
x^2 & \text{if } -2 \leq x < 0 \\
3x & \text{if } 0 < x < 3 \\
\frac{x}{x-4} & \text{if } 3 \leq x \leq 6 \\
\frac{x}{x-5} & \text{if } 6 < x \leq 7 \\
x^2 & \text{if } x > 7 
\end{cases}
\]

We need to check for discontinuities at the boundaries of each interval, as well as within the intervals where the function might not be defined.

### Step 1: Check at \( x = -2 \)
- Left-hand limit: \(\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} -2x = -2(-2) = 4\)
- Right-hand limit: \(\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} x^2 = (-2)^2 = 4\)
- Function value: \(f(-2) = (-2)^2 = 4\)

Since the left-hand limit, right-hand limit, and function value are all equal at \( x = -2 \), there is **no discontinuity** at this point.

### Step 2: Check at \( x = 0 \)
- Left-hand limit: \(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0^2 = 0\)
- Right-hand limit: \(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3x = 3(0) = 0\)

Since the left-hand and right-hand limits are both 0, and the function is not defined at \( x = 0 \) (the intervals are open at \( x = 0 \)), there is **no discontinuity** at this point.

### Step 3: Check at \( x = 3 \)
- Left-hand limit: \(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 3x = 3(3) = 9\)
- Right-hand limit: \(\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x}{x-4} = \frac{3}{3-4} = \frac{3}{-1} = -3\)
- Function value: \(f(3) = \frac{3}{3-4} = -3\)

The left-hand limit and right-hand limit are not equal at \( x = 3 \), so there is a **discontinuity** at \( x = 3 \).

### Step 4: Check at \( x = 4 \)
- Within the interval \(3 \leq x \leq 6\), the function \(f(x) = \frac{x}{x-4}\) has a vertical asymptote at \(x = 4\), making it undefined at this point.

Therefore, there is a **discontinuity** at \( x = 4 \).

### Step 5: Check at \( x = 6 \)
- Left-hand limit: \(\lim_{x \to 6^-} f(x) = \lim_{x \to 6^-} \frac{x}{x-4} = \frac{6}{6-4} = \frac{6}{2} = 3\)
- Right-hand limit: \(\lim_{x \to 6^+} f(x) = \lim_{x \to 6^+} \frac{x}{x-5} = \frac{6}{6-5} = \frac{6}{1} = 6\)
- Function value: \(f(6) = \frac{6}{6-4} = 3\)

The left-hand limit and right-hand limit are not equal at \( x = 6 \), so there is a **discontinuity** at \( x = 6 \).

### Step 6: Check at \( x = 7 \)
- Left-hand limit: \(\lim_{x \to 7^-} f(x) = \lim_{x \to 7^-} \frac{x}{x-5} = \frac{7}{7-5} = \frac{7}{2} = 3.5\)
- Right-hand limit: \(\lim_{x \to 7^+} f(x) = \lim_{x \to 7^+} x^2 = 7^2 = 49\)
- Function value: \(f(7) = \frac{7}{7-5} = 3.5\)

The left-hand limit and right-hand limit are not equal at \( x = 7 \), so there is a **discontinuity** at \( x = 7 \).

### Conclusion
The points of discontinuity for the function \( f(x) \) are at \( x = 3 \), \( x = 4 \), \( x = 6 \), and \( x = 7 \).

Would you like a detailed explanation of any part? Here are some related questions:

1. How do you determine the type of discontinuity (jump, infinite, or removable)?
2. What is the difference between left-hand and right-hand limits?
3. How does the definition of a piecewise function affect continuity?
4. Can a function have an infinite number of discontinuities?
5. How do you analyze the continuity of more complex piecewise functions?

**Tip:** When checking continuity, always consider both limits (left-hand and right-hand) and the function's actual value at the point of interest.

Explore how to find points of discontinuity for the piecewise function f(x) = -2x, x^2, 3x, x/(x-4), x/(x-5), x^2 in detail. Learn about left-hand limits, right-hand limits, and function values at critical points.

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