# Find the points of local maxima and local minima of f(x) = x 3 – 6 x 2 + 9x + 2014, x ε �

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To find the local maxima and minima of the function \( f(x) = x^3 - 6x^2 + 9x + 2014 \), we need to follow these steps:

1. Find the first derivative \( f'(x) \).

2. Set \( f'(x) = 0 \) to find critical points.

3. Find the second derivative \( f''(x) \).

4. Use the second derivative test to classify the critical points as local maxima, minima, or points of inflection.

Let's go through the steps:

1. Find \( f'(x) \):

\[ f'(x) = 3x^2 - 12x + 9 \]

2. Set \( f'(x) = 0 \) to find critical points:

\[ 3x^2 - 12x + 9 = 0 \]

\[ (x - 3)^2 = 0 \]

Solve for \( x \):

\[ x - 3 = 0 \]

\[ x = 3 \]

So, the critical point is \( x = 3 \).

3. Find \( f''(x) \):

\[ f''(x) = 6x - 12 \]

4. Use the second derivative test:

- When \( f''(x) > 0 \), the function is concave up, indicating a local minimum.

- When \( f''(x) < 0 \), the function is concave down, indicating a local maximum.

Evaluate \( f''(3) \):

\[ f''(3) = 6(3) - 12 = 6 \]

Since \( f''(3) > 0 \), the critical point \( x = 3 \) is a local minimum.

So, the function \( f(x) = x^3 - 6x^2 + 9x + 2014 \) has a local minimum at \( x = 3 \).