Examine continuity of the function `f(x) = x^(3) + 2x^(2)- 1` at `x = 1`.

Explanation

To examine the continuity of the function \( f(x) = x^3 + 2x^2 - 1 \) at \( x = 1 \), we need to check the following conditions:

1. The function \( f(x) \) must be defined at \( x = 1 \).
2. The left-hand limit as \( x \) approaches 1 must exist.
3. The right-hand limit as \( x \) approaches 1 must exist.
4. The left-hand limit and the right-hand limit must be equal to each other and also equal to \( f(1) \).

Let's go through these steps one by one.

### Step 1: Check if \( f(1) \) is defined

First, we calculate \( f(1) \):
\[ f(1) = 1^3 + 2(1^2) - 1 = 1 + 2 - 1 = 2 \]
So, \( f(1) \) is defined and equals 2.

### Step 2: Calculate the left-hand limit as \( x \) approaches 1

The left-hand limit is defined as:
\[ \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) \]
Substituting \( f(x) \) into the limit:
\[ \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} ((1 - h)^3 + 2(1 - h)^2 - 1) \]
Now, we expand \( (1 - h)^3 \) and \( (1 - h)^2 \):
\[ (1 - h)^3 = 1 - 3h + 3h^2 - h^3 \]
\[ (1 - h)^2 = 1 - 2h + h^2 \]
Substituting these expansions back into the limit:
\[ \lim_{h \to 0} \left( (1 - 3h + 3h^2 - h^3) + 2(1 - 2h + h^2) - 1 \right) \]
Simplifying this expression:
\[ = \lim_{h \to 0} \left( 1 - 3h + 3h^2 - h^3 + 2 - 4h + 2h^2 - 1 \right) \]
\[ = \lim_{h \to 0} \left( 2 - 7h + 5h^2 - h^3 \right) \]
Taking the limit as \( h \to 0 \):
\[ = 2 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1

The right-hand limit is defined as:
\[ \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) \]
Substituting \( f(x) \) into the limit:
\[ \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} ((1 + h)^3 + 2(1 + h)^2 - 1) \]
Now, we expand \( (1 + h)^3 \) and \( (1 + h)^2 \):
\[ (1 + h)^3 = 1 + 3h + 3h^2 + h^3 \]
\[ (1 + h)^2 = 1 + 2h + h^2 \]
Substituting these expansions back into the limit:
\[ \lim_{h \to 0} \left( (1 + 3h + 3h^2 + h^3) + 2(1 + 2h + h^2) - 1 \right) \]
Simplifying this expression:
\[ = \lim_{h \to 0} \left( 1 + 3h + 3h^2 + h^3 + 2 + 4h + 2h^2 - 1 \right) \]
\[ = \lim_{h \to 0} \left( 2 + 7h + 5h^2 + h^3 \right) \]
Taking the limit as \( h \to 0 \):
\[ = 2 \]

### Step 4: Compare the left-hand limit, right-hand limit, and \( f(1) \)

We have:
- Left-hand limit: \( 2 \)
- Right-hand limit: \( 2 \)
- \( f(1) = 2 \)

Since the left-hand limit equals the right-hand limit and both are equal to \( f(1) \), we conclude that the function \( f(x) \) is continuous at \( x = 1 \).

### Conclusion

The function \( f(x) = x^3 + 2x^2 - 1 \) is continuous at \( x = 1 \).

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