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

Answer

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

### Step 1: Find the first derivative of the function
To find the critical points, we first need to compute the first derivative of the function \( f(x) \).

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

Using the power rule:

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

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

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

### Step 3: Simplify the equation
We can simplify this equation by dividing all terms by 6:

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

### Step 4: Factor the quadratic equation
Now, we will factor the quadratic equation:

\[ (x - 1)(x - 6) = 0 \]

### Step 5: Solve for \( x \)
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 6: Determine the nature of the critical points
To determine whether these points are local maxima or minima, we will use the second derivative test. First, we find the second derivative of \( f(x) \).

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

Using the power rule again:

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

Now we 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 local maximum.

2. For \( x = 6 \):
\[ f''(6) = 12(6) - 42 = 72 - 42 = 30 \]
Since \( f''(6) > 0 \), this indicates that \( x = 6 \) is a local minimum.

### Conclusion
The points of local maxima and minima are:
- Local maximum at \( x = 1 \)
- Local minimum at \( x = 6 \)