Find the maximum and minimum values of the function :
`f(x)=2x^(3)-15x^(2)+36x+11.`

Answer

To find the maximum and minimum values of the function \( f(x) = 2x^3 - 15x^2 + 36x + 11 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(2x^3) - \frac{d}{dx}(15x^2) + \frac{d}{dx}(36x) + \frac{d}{dx}(11) \]

Calculating the derivatives:

\[ f'(x) = 6x^2 - 30x + 36 \]

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

\[ 6x^2 - 30x + 36 = 0 \]

Dividing the entire equation by 6 simplifies it:

\[ x^2 - 5x + 6 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:

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

### Step 4: Solve for critical points
Setting each factor to zero gives us the critical points:

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 5: Find the second derivative
To determine whether these critical points are maxima or minima, we find the second derivative:

\[ f''(x) = \frac{d}{dx}(6x^2 - 30x + 36) = 12x - 30 \]

### Step 6: Evaluate the second derivative at critical points
Now we evaluate \( f''(x) \) at the critical points \( x = 2 \) and \( x = 3 \):

1. For \( x = 2 \):

\[ f''(2) = 12(2) - 30 = 24 - 30 = -6 \quad (\text{Negative, hence local maximum}) \]

2. For \( x = 3 \):

\[ f''(3) = 12(3) - 30 = 36 - 30 = 6 \quad (\text{Positive, hence local minimum}) \]

### Step 7: Calculate the function values at critical points
Now we calculate the function values at these critical points to find the maximum and minimum values.

1. For \( x = 2 \):

\[ f(2) = 2(2^3) - 15(2^2) + 36(2) + 11 \]
\[ = 2(8) - 15(4) + 72 + 11 \]
\[ = 16 - 60 + 72 + 11 = 39 \]

2. For \( x = 3 \):

\[ f(3) = 2(3^3) - 15(3^2) + 36(3) + 11 \]
\[ = 2(27) - 15(9) + 108 + 11 \]
\[ = 54 - 135 + 108 + 11 = 38 \]

### Conclusion
Thus, the maximum value of the function is \( f(2) = 39 \) at \( x = 2 \) and the minimum value is \( f(3) = 38 \) at \( x = 3 \).