Determine why the function f(x)=⎧⎩⎨⎪⎪3−x,x, 2x2,x<5x=5x>5fx={3−x,x<5x, x=52x2,x>5 is discontinuous at a given point x=5x=5 on a graph. State the condition which fails first.

To determine why the function 

\[
f(x) = \begin{cases} 
3 - x, & x < 5 \\ 
x, & x = 5 \\ 
2x^2, & x > 5 
\end{cases}
\]

is discontinuous at \( x = 5 \), we need to check if the function satisfies the conditions for continuity at this point. A function \( f(x) \) is continuous at \( x = 5 \) if it meets all three of the following conditions:

1. **\( f(5) \) is defined.**
2. **\( \lim_{x \to 5} f(x) \) exists.**
3. **\( \lim_{x \to 5} f(x) = f(5) \).**

Let's examine each condition step-by-step.

### Step 1: Check if \( f(5) \) is defined
According to the piecewise definition, at \( x = 5 \), we use the middle part of the function:
\[
f(5) = 5.
\]
So, \( f(5) \) is defined.

### Step 2: Find \( \lim_{x \to 5} f(x) \)
To find \( \lim_{x \to 5} f(x) \), we need to check both the left-hand limit (\( \lim_{x \to 5^-} f(x) \)) and the right-hand limit (\( \lim_{x \to 5^+} f(x) \)).

#### Left-hand limit (\( x \to 5^- \))
When \( x \to 5 \) from the left (\( x < 5 \)), \( f(x) = 3 - x \).
\[
\lim_{x \to 5^-} f(x) = 3 - 5 = -2.
\]

#### Right-hand limit (\( x \to 5^+ \))
When \( x \to 5 \) from the right (\( x > 5 \)), \( f(x) = 2x^2 \).
\[
\lim_{x \to 5^+} f(x) = 2 \cdot (5)^2 = 2 \cdot 25 = 50.
\]

Since \( \lim_{x \to 5^-} f(x) = -2 \) and \( \lim_{x \to 5^+} f(x) = 50 \), the left-hand and right-hand limits do not match. Therefore, \( \lim_{x \to 5} f(x) \) does not exist.

### Step 3: Conclusion on Continuity
Since the limit \( \lim_{x \to 5} f(x) \) does not exist, the function \( f(x) \) fails the second condition for continuity. Therefore, \( f(x) \) is **discontinuous at \( x = 5 \)**. The condition that fails first is the existence of \( \lim_{x \to 5} f(x) \), as the left and right limits are not equal.

Would you like further details or have any questions?

---

Here are 5 questions to deepen understanding:
1. What would change if both left-hand and right-hand limits were equal at \( x = 5 \)?
2. How would the function's graph look around \( x = 5 \) given the discontinuity?
3. How do you find the points of discontinuity in more complex piecewise functions?
4. What type of discontinuity is present at \( x = 5 \) in this function?
5. How can we modify \( f(x) \) to make it continuous at \( x = 5 \)?

**Tip:** Always check the limits from both sides when assessing continuity for piecewise functions!

Determine why the function f(x) = {3 - x, x < 5; x, x = 5; 2x^2, x > 5} is discontinuous at a given point x = 5 on a graph. State the condition which fails first.

Explore why the piecewise function f(x) = {3 - x for x < 5; x for x = 5; 2x^2 for x > 5} is discontinuous at x = 5. This example covers continuity criteria, limit calculations, and condition verification to identify discontinuity at x = 5, suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation