Let f(x) be a polynomial of degree 3 such that `f(-2)=5, f(2)=-3,`` f'(x)` has a critical point at `x = -2` and `f''(x)` has a critical point at x = 2. Then f(x) has a local maxima at x = a and local minimum at x = b. Then find b-a.

Explanation

To solve the problem step by step, we will analyze the given conditions and derive the necessary equations.

### Step 1: Understanding the conditions
We know that \( f(x) \) is a polynomial of degree 3. The conditions given are:
- \( f(-2) = 5 \)
- \( f(2) = -3 \)
- \( f'(x) \) has a critical point at \( x = -2 \)
- \( f''(x) \) has a critical point at \( x = 2 \)

### Step 2: Finding \( f'(x) \)
Since \( f(x) \) is a cubic polynomial, we can express it as:
\[ f(x) = ax^3 + bx^2 + cx + d \]
Differentiating, we get:
\[ f'(x) = 3ax^2 + 2bx + c \]
Given that \( f'(-2) = 0 \), we substitute \( x = -2 \):
\[ f'(-2) = 3a(-2)^2 + 2b(-2) + c = 0 \]
This simplifies to:
\[ 12a - 4b + c = 0 \quad \text{(1)} \]

### Step 3: Finding \( f''(x) \)
Differentiating \( f'(x) \):
\[ f''(x) = 6ax + 2b \]
Since \( f''(2) = 0 \), we substitute \( x = 2 \):
\[ f''(2) = 6a(2) + 2b = 0 \]
This simplifies to:
\[ 12a + 2b = 0 \quad \text{(2)} \]

### Step 4: Solving the system of equations
From equation (2), we can express \( b \) in terms of \( a \):
\[ b = -6a \]
Substituting \( b \) into equation (1):
\[ 12a - 4(-6a) + c = 0 \]
This simplifies to:
\[ 12a + 24a + c = 0 \implies 36a + c = 0 \implies c = -36a \quad \text{(3)} \]

### Step 5: Using the values of \( f(-2) \) and \( f(2) \)
Now we use the conditions \( f(-2) = 5 \) and \( f(2) = -3 \).

Substituting \( x = -2 \) into \( f(x) \):
\[ f(-2) = a(-2)^3 + b(-2)^2 + c(-2) + d = 5 \]
Substituting \( b \) and \( c \):
\[ -8a + 4(-6a) + 72a + d = 5 \implies -8a - 24a + 72a + d = 5 \implies 40a + d = 5 \quad \text{(4)} \]

Substituting \( x = 2 \) into \( f(x) \):
\[ f(2) = a(2)^3 + b(2)^2 + c(2) + d = -3 \]
Substituting \( b \) and \( c \):
\[ 8a + 4(-6a) + (-72a) + d = -3 \implies 8a - 24a - 72a + d = -3 \implies -88a + d = -3 \quad \text{(5)} \]

### Step 6: Solving equations (4) and (5)
From equations (4) and (5):
1. \( 40a + d = 5 \)
2. \( -88a + d = -3 \)

Subtracting these equations:
\[ (40a + d) - (-88a + d) = 5 - (-3) \]
This simplifies to:
\[ 128a = 8 \implies a = \frac{1}{16} \]

Substituting \( a \) back into equation (4):
\[ 40\left(\frac{1}{16}\right) + d = 5 \implies \frac{40}{16} + d = 5 \implies \frac{5}{2} + d = 5 \implies d = 5 - \frac{5}{2} = \frac{5}{2} \]

Now substituting \( a \) into equation (3) to find \( c \):
\[ c = -36\left(\frac{1}{16}\right) = -\frac{36}{16} = -\frac{9}{4} \]

And substituting \( a \) into \( b \):
\[ b = -6\left(\frac{1}{16}\right) = -\frac{6}{16} = -\frac{3}{8} \]

### Step 7: Finding critical points
Now we have:
- \( a = \frac{1}{16} \)
- \( b = -\frac{3}{8} \)
- \( c = -\frac{9}{4} \)
- \( d = \frac{5}{2} \)

The polynomial \( f(x) \) can be expressed as:
\[ f(x) = \frac{1}{16}x^3 - \frac{3}{8}x^2 - \frac{9}{4}x + \frac{5}{2} \]

### Step 8: Finding local maxima and minima
The critical points from \( f'(x) = 0 \):
\[ f'(x) = \frac{1}{16}(3x^2 - 6x - 72) = 0 \]
Factoring gives critical points at \( x = -2 \) and \( x = 6 \).

### Step 9: Determining local maxima and minima
To determine which is a maximum and which is a minimum, we can evaluate \( f''(x) \):
- \( f''(-2) < 0 \) indicates a local maximum at \( x = -2 \)
- \( f''(6) > 0 \) indicates a local minimum at \( x = 6 \)

### Step 10: Finding \( b - a \)
Let \( a = -2 \) (local maximum) and \( b = 6 \) (local minimum):
\[ b - a = 6 - (-2) = 6 + 2 = 8 \]

### Final Answer
\[ \boxed{8} \]