If the function `f(x)=2x^3-6x^2-18x` has local maxima at `x=a` and local minima at `x=b` and the area bounded by `y=f(x)` from `x=a` to `x=b` is A. Then find the value of `4A`

Answer

To solve the problem step by step, we will follow these main steps:

1. **Find the critical points of the function** by taking the first derivative and setting it to zero.
2. **Determine the nature of the critical points** using the second derivative test.
3. **Calculate the area between the curve and the x-axis** from the local maxima to the local minima.
4. **Compute the final value of \(4A\)**.

### Step 1: Find the critical points

Given the function:
\[ f(x) = 2x^3 - 6x^2 - 18x \]

First, we differentiate \(f(x)\):
\[ f'(x) = \frac{d}{dx}(2x^3 - 6x^2 - 18x) = 6x^2 - 12x - 18 \]

Now, set the first derivative equal to zero to find critical points:
\[ 6x^2 - 12x - 18 = 0 \]
Dividing the entire equation by 6:
\[ x^2 - 2x - 3 = 0 \]

Factoring the quadratic:
\[ (x - 3)(x + 1) = 0 \]
Thus, the critical points are:
\[ x = 3 \quad \text{and} \quad x = -1 \]

### Step 2: Determine the nature of the critical points

Next, we find the second derivative:
\[ f''(x) = \frac{d}{dx}(6x^2 - 12x - 18) = 12x - 12 \]

Now we evaluate the second derivative at the critical points:
- For \(x = 3\):
\[ f''(3) = 12(3) - 12 = 36 > 0 \]
This indicates a local minimum at \(x = 3\).

- For \(x = -1\):
\[ f''(-1) = 12(-1) - 12 = -24 < 0 \]
This indicates a local maximum at \(x = -1\).

Thus, we have:
- Local maxima at \(x = -1\) (denote \(a = -1\))
- Local minima at \(x = 3\) (denote \(b = 3\))

### Step 3: Calculate the area between the curve and the x-axis

The area \(A\) bounded by the curve from \(x = a\) to \(x = b\) is given by:
\[ A = \int_{-1}^{3} f(x) \, dx \]

We can break this integral into two parts:
\[ A = \int_{-1}^{0} f(x) \, dx + \int_{0}^{3} f(x) \, dx \]

Calculating the first integral:
\[ \int_{-1}^{0} (2x^3 - 6x^2 - 18x) \, dx \]
Calculating the second integral:
\[ \int_{0}^{3} (2x^3 - 6x^2 - 18x) \, dx \]

Now, we calculate the integrals:
1. **For \( \int (2x^3 - 6x^2 - 18x) \, dx \)**:
\[ = \frac{2}{4}x^4 - \frac{6}{3}x^3 - \frac{18}{2}x^2 = \frac{1}{2}x^4 - 2x^3 - 9x^2 \]

2. **Evaluating from \(-1\) to \(0\)**:
\[ \left[ \frac{1}{2}(0)^4 - 2(0)^3 - 9(0)^2 \right] - \left[ \frac{1}{2}(-1)^4 - 2(-1)^3 - 9(-1)^2 \right] \]
\[ = 0 - \left[ \frac{1}{2} - 2 - 9 \right] = 0 - \left[ \frac{1}{2} + 2 + 9 \right] = 0 - \frac{19}{2} = -\frac{19}{2} \]

3. **Evaluating from \(0\) to \(3\)**:
\[ \left[ \frac{1}{2}(3)^4 - 2(3)^3 - 9(3)^2 \right] - \left[ \frac{1}{2}(0)^4 - 2(0)^3 - 9(0)^2 \right] \]
\[ = \left[ \frac{1}{2}(81) - 54 - 81 \right] - 0 = \frac{81}{2} - 54 - 81 = \frac{81}{2} - \frac{108}{2} - \frac{162}{2} = \frac{81 - 108 - 162}{2} = -\frac{189}{2} \]

### Step 4: Combine the areas

The total area \(A\) is:
\[ A = \left| -\frac{19}{2} \right| + \left| -\frac{189}{2} \right| = \frac{19}{2} + \frac{189}{2} = \frac{208}{2} = 104 \]

Finally, we need to find \(4A\):
\[ 4A = 4 \times 104 = 416 \]

### Final Answer
Thus, the value of \(4A\) is:
\[ \boxed{416} \]