Apply the definition of continuity to determine if a function is continuous at x = a.

Consider the following function:

f left parenthesis x right parenthesis equals open curly brackets table attributes columnalign left columnspacing 1.4ex end attributes row cell square root of x squared plus 7 end root end cell cell i f space x less or equal than 3 end cell row cell left parenthesis x minus 1 right parenthesis squared end cell cell i f space x greater than 3 end cell end table close

Determine if f left parenthesis x right parenthesis is continuous at x equals 3. If not, select the option with the correct reasoning as to why not.

To determine whether the given function \( f(x) \) is continuous at \( x = 3 \), we need to apply the definition of continuity.

A function \( f(x) \) is continuous at \( x = a \) if the following three conditions are satisfied:

1. **\( f(a) \) is defined**: The function must be defined at \( x = a \).
2. **\( \lim_{x \to a} f(x) \) exists**: The limit of the function as \( x \) approaches \( a \) from both sides must exist.
3. **\( \lim_{x \to a} f(x) = f(a) \)**: The value of the limit as \( x \) approaches \( a \) must equal the value of the function at \( x = a \).

Let's check these conditions at \( x = 3 \).

### 1. Check if \( f(3) \) is defined:
The function \( f(x) \) is defined piecewise:
- For \( x \leq 3 \), \( f(x) = \sqrt{x^2 + 7} \).
- For \( x > 3 \), \( f(x) = (x - 1)^2 \).

Since \( 3 \leq 3 \), we use the first piece of the function to find \( f(3) \):
\[
f(3) = \sqrt{3^2 + 7} = \sqrt{9 + 7} = \sqrt{16} = 4.
\]
Thus, \( f(3) = 4 \), so the function is defined at \( x = 3 \).

### 2. Check if \( \lim_{x \to 3} f(x) \) exists:
We need to check the left-hand limit (\( \lim_{x \to 3^-} f(x) \)) and the right-hand limit (\( \lim_{x \to 3^+} f(x) \)) to ensure that both exist and are equal.

#### Left-hand limit (\( \lim_{x \to 3^-} f(x) \)):
For \( x \leq 3 \), \( f(x) = \sqrt{x^2 + 7} \). So, we calculate the limit as \( x \) approaches 3 from the left:
\[
\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \sqrt{x^2 + 7} = \sqrt{3^2 + 7} = \sqrt{16} = 4.
\]

#### Right-hand limit (\( \lim_{x \to 3^+} f(x) \)):
For \( x > 3 \), \( f(x) = (x - 1)^2 \). So, we calculate the limit as \( x \) approaches 3 from the right:
\[
\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (x - 1)^2 = (3 - 1)^2 = 2^2 = 4.
\]

Since both the left-hand limit and right-hand limit are equal to 4, the overall limit exists and is equal to 4:
\[
\lim_{x \to 3} f(x) = 4.
\]

### 3. Check if \( \lim_{x \to 3} f(x) = f(3) \):
We have:
- \( \lim_{x \to 3} f(x) = 4 \),
- \( f(3) = 4 \).

Since these values are equal, the third condition is satisfied.

### Conclusion:
Since all three conditions for continuity are satisfied, the function \( f(x) \) is **continuous at \( x = 3 \)**.

Would you like more details or have any questions on the steps? Here are 5 related questions:

1. How would the continuity be affected if the function definition changed for \( x = 3 \)?
2. Can a function be continuous if its left-hand limit and right-hand limit are not equal?
3. What happens to continuity if \( f(a) \) is not defined?
4. How can you determine the continuity of more complex piecewise functions?
5. Is it possible for a function to be discontinuous at a point, but still have limits from both sides?

**Tip**: When checking continuity at a point, always verify the function's value, the limit's existence, and the equality between them to ensure the function is continuous.

Apply the definition of continuity to determine if a function is continuous at x = a. Consider the following function: f(x) = {sqrt(x^2 + 7), if x <= 3; (x - 1)^2, if x > 3}. Determine if f(x) is continuous at x = 3. If not, select the option with the correct reasoning as to why not.

Analyze the continuity of the function f(x), defined as a piecewise function with sqrt(x^2 + 7) for x <= 3 and (x - 1)^2 for x > 3, at x = 3. This problem demonstrates how to apply the definition of continuity, check limits from both sides, and verify function values. Ideal for students in Grades 11-12 studying advanced calculus.

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