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 need to check if the following condition holds:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

### Step 1: Calculate \( f(1) \)

First, we calculate the value of the function at \( x = 1 \):

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

### Step 2: Calculate the Left-Hand Limit \( \lim_{x \to 1^-} f(x) \)

Next, we calculate the left-hand limit as \( x \) approaches 1:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^3 + 2x^2 - 1) \]

Substituting \( x = 1 - h \) (where \( h \to 0^+ \)):

\[ \lim_{h \to 0} f(1 - h) = \lim_{h \to 0} ((1 - h)^3 + 2(1 - h)^2 - 1) \]

Calculating \( (1 - h)^3 \) and \( (1 - h)^2 \):

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

Thus,

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

Simplifying this:

\[ = 1 - 3h + 3h^2 - h^3 + 2 - 4h + 2h^2 - 1 \]
\[ = 2 - 7h + 5h^2 - h^3 \]

Taking the limit as \( h \to 0 \):

\[ \lim_{h \to 0} (2 - 7h + 5h^2 - h^3) = 2 \]

So,

\[ \lim_{x \to 1^-} f(x) = 2 \]

### Step 3: Calculate the Right-Hand Limit \( \lim_{x \to 1^+} f(x) \)

Now, we calculate the right-hand limit as \( x \) approaches 1:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^3 + 2x^2 - 1) \]

Substituting \( x = 1 + h \) (where \( h \to 0^+ \)):

\[ \lim_{h \to 0} f(1 + h) = \lim_{h \to 0} ((1 + h)^3 + 2(1 + h)^2 - 1) \]

Calculating \( (1 + h)^3 \) and \( (1 + h)^2 \):

\[ (1 + h)^3 = 1 + 3h + 3h^2 + h^3 \]
\[ (1 + h)^2 = 1 + 2h + h^2 \]

Thus,

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

Simplifying this:

\[ = 1 + 3h + 3h^2 + h^3 + 2 + 4h + 2h^2 - 1 \]
\[ = 2 + 7h + 5h^2 + h^3 \]

Taking the limit as \( h \to 0 \):

\[ \lim_{h \to 0} (2 + 7h + 5h^2 + h^3) = 2 \]

So,

\[ \lim_{x \to 1^+} f(x) = 2 \]

### Step 4: Conclusion

Since we have:

\[ \lim_{x \to 1^-} f(x) = 2, \quad \lim_{x \to 1^+} f(x) = 2, \quad \text{and} \quad f(1) = 2 \]

We conclude that:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Thus, the function \( f(x) \) is continuous at \( x = 1 \).