If `f(x)=2x^(3)-21x^(2)+36x-20`, then

Options

A. f has maxima at x=6
B. f has minima at x=6
C. f has maximum value -128
D. f has minimum value -3

Answer

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

### Step 1: Find the first derivative \( f'(x) \)

The first derivative of the function will help us find the critical points where the function may have maximum or minimum values.

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

Using the power rule of differentiation:

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

### Step 2: Set the first derivative equal to zero

To find the critical points, we set \( f'(x) = 0 \):

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

### Step 3: Simplify the equation

We can simplify the equation by dividing everything by 6:

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

### Step 4: Factor the quadratic equation

Now we will factor the quadratic:

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

### Step 5: Solve for \( x \)

Setting each factor to zero gives us the critical points:

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

### Step 6: Find the second derivative \( f''(x) \)

To determine whether these critical points are maxima or minima, we will find the second derivative:

\[ f''(x) = \frac{d}{dx}(6x^2 - 42x + 36) = 12x - 42 \]

### Step 7: Evaluate the second derivative at the critical points

Now we will evaluate \( f''(x) \) at the critical points \( x = 1 \) and \( x = 6 \).

1. For \( x = 1 \):

\[ f''(1) = 12(1) - 42 = 12 - 42 = -30 \quad (\text{less than } 0 \Rightarrow \text{maximum}) \]

2. For \( x = 6 \):

\[ f''(6) = 12(6) - 42 = 72 - 42 = 30 \quad (\text{greater than } 0 \Rightarrow \text{minimum}) \]

### Step 8: Find the maximum and minimum values

Now we will evaluate the function \( f(x) \) at the critical points to find the maximum and minimum values.

1. For \( x = 1 \):

\[ f(1) = 2(1)^3 - 21(1)^2 + 36(1) - 20 = 2 - 21 + 36 - 20 = -3 \]

2. For \( x = 6 \):

\[ f(6) = 2(6)^3 - 21(6)^2 + 36(6) - 20 = 2(216) - 21(36) + 216 - 20 \]
Calculating each term:
\[ = 432 - 756 + 216 - 20 = -128 \]

### Conclusion

- The maximum value of \( f(x) \) occurs at \( x = 1 \) and is \( -3 \).
- The minimum value of \( f(x) \) occurs at \( x = 6 \) and is \( -128 \).