If `f(x)=2x^(3)-21x^(2)+36x-30`, then which one of the following is correct?

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 of minima

Explanation

To determine the nature of the function \( f(x) = 2x^3 - 21x^2 + 36x - 30 \), we need to find its critical points and analyze them using the first and second derivative tests.

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

The first derivative of \( f(x) \) is calculated as follows:

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

Calculating each term:

- \( \frac{d}{dx}(2x^3) = 6x^2 \)
- \( \frac{d}{dx}(21x^2) = 42x \)
- \( \frac{d}{dx}(36x) = 36 \)
- \( \frac{d}{dx}(30) = 0 \)

Thus, we have:

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

### Step 2: Set the first derivative to zero to find critical points

Now, we set \( f'(x) = 0 \):

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

Dividing the entire equation by 6:

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

### Step 3: Factor the quadratic equation

We can factor the quadratic:

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

Thus, the critical points are:

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

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

Next, we find the second derivative:

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

Calculating each term:

- \( \frac{d}{dx}(6x^2) = 12x \)
- \( \frac{d}{dx}(-42x) = -42 \)
- \( \frac{d}{dx}(36) = 0 \)

Thus, we have:

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

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

Now, we evaluate \( f''(x) \) at the critical points:

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

Based on the analysis, we find:
- \( f(x) \) has a local maximum at \( x = 1 \).
- \( f(x) \) has a local minimum at \( x = 6 \).

Thus, the correct option is that \( f(x) \) has a maximum at \( x = 1 \).