Examine contnuity 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 will follow these steps:

### Step 1: Determine \( f(1) \)
First, we need to evaluate the function at \( x = 1 \).

\[ f(1) = 1^3 + 2(1^2) - 1 \]
\[ = 1 + 2 - 1 \]
\[ = 2 \]

### Step 2: Calculate the Left-Hand Limit as \( x \) approaches 1
Next, we calculate the left-hand limit of \( f(x) \) as \( x \) approaches 1.

\[ \lim_{x \to 1^-} f(x) = \lim_{h \to 0} f(1 - h) = f(1 - h) \]
Substituting \( 1 - h \) into the function:
\[ f(1 - h) = (1 - h)^3 + 2(1 - h)^2 - 1 \]
Expanding this:
\[ = (1 - 3h + 3h^2 - h^3) + 2(1 - 2h + h^2) - 1 \]
\[ = 1 - 3h + 3h^2 - h^3 + 2 - 4h + 2h^2 - 1 \]
Combining like terms:
\[ = (1 + 2 - 1) + (-3h - 4h) + (3h^2 + 2h^2) - h^3 \]
\[ = 2 - 7h + 5h^2 - h^3 \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} (2 - 7h + 5h^2 - h^3) = 2 \]

### Step 3: Calculate the Right-Hand Limit as \( x \) approaches 1
Now, we calculate the right-hand limit of \( f(x) \) as \( x \) approaches 1.

\[ \lim_{x \to 1^+} f(x) = \lim_{h \to 0} f(1 + h) = f(1 + h) \]
Substituting \( 1 + h \) into the function:
\[ f(1 + h) = (1 + h)^3 + 2(1 + h)^2 - 1 \]
Expanding this:
\[ = (1 + 3h + 3h^2 + h^3) + 2(1 + 2h + h^2) - 1 \]
\[ = 1 + 3h + 3h^2 + h^3 + 2 + 4h + 2h^2 - 1 \]
Combining like terms:
\[ = (1 + 2 - 1) + (3h + 4h) + (3h^2 + 2h^2) + h^3 \]
\[ = 2 + 7h + 5h^2 + h^3 \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} (2 + 7h + 5h^2 + h^3) = 2 \]

### Step 4: Compare the Limits and the Function Value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 1 \):

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

Since all three values are equal, we conclude that the function is continuous at \( x = 1 \).

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

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