Find the local maxima and minima of the function `f(x)=2x^(3)-21x^(2)+36x-20`.

Answer

To find the local maxima and minima of the function \( f(x) = 2x^3 - 21x^2 + 36x - 20 \), we will follow these steps:

### Step 1: Find the first derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x - 20) \]

Using the power rule, we get:

\[ f'(x) = 6x^2 - 42x + 36 \]

### Step 2: Set the first derivative to zero to find critical points
Next, we set the first derivative equal to zero to find the critical points.

\[ 6x^2 - 42x + 36 = 0 \]

We can simplify this equation by dividing everything by 6:

\[ x^2 - 7x + 6 = 0 \]

### Step 3: Factor the quadratic equation
Now we will factor the quadratic equation:

\[ x^2 - 7x + 6 = (x - 1)(x - 6) = 0 \]

Setting each factor to zero gives us the critical points:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 6 = 0 \quad \Rightarrow \quad x = 6 \]

### Step 4: Find the second derivative of the function
To determine whether these critical points are maxima or minima, we need to find the second derivative of \( f(x) \).

\[ f''(x) = \frac{d}{dx}(6x^2 - 42x + 36) \]

Calculating the second derivative, we get:

\[ f''(x) = 12x - 42 \]

### Step 5: Evaluate the second derivative at the critical points
Now we will evaluate the second derivative at the critical points \( x = 1 \) and \( x = 6 \).

1. For \( x = 1 \):

\[ f''(1) = 12(1) - 42 = 12 - 42 = -30 \]

Since \( f''(1) < 0 \), this indicates that \( x = 1 \) is a point of local maxima.

2. For \( x = 6 \):

\[ f''(6) = 12(6) - 42 = 72 - 42 = 30 \]

Since \( f''(6) > 0 \), this indicates that \( x = 6 \) is a point of local minima.

### Conclusion
Thus, the function has:
- A local maximum at \( x = 1 \)
- A local minimum at \( x = 6 \)