If `f(x) = 2 x ^(3) - 21 x^(2) + 36 x - 30`, then

Options

A. f(x) has minimum at x = 1
B. f(x) has maximum at x = 6
C. f(x) has maximum at x = 1
D. f(x) has no maxima or minima

Answer

To determine whether the function \( f(x) = 2x^3 - 21x^2 + 36x - 30 \) has local maxima or minima at specific points, we will follow these steps:

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

To find the critical points, we first need to differentiate the function:

\[ f'(x) = \frac{d}{dx}(2x^3 - 21x^2 + 36x - 30) \]

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:

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

Dividing the entire equation by 6 simplifies it:

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

### Step 3: Factor the quadratic equation

Now we factor the quadratic equation:

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

Setting each factor to zero gives us the critical points:

\[ x = 1 \quad \text{and} \quad x = 6 \]

### Step 4: Find the second derivative \( f''(x) \)

Now we will find the second derivative to determine the nature of the critical points:

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

Using the power rule again:

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

### Step 5: Evaluate the second derivative at the critical points

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 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, we conclude that:
- \( x = 1 \) is a local maximum.
- \( x = 6 \) is a local minimum.

### Final Answer

- Maximum at \( x = 1 \)
- Minimum at \( x = 6 \)

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