Let `g(x)=a_(0)+a_(1)x+a_(2)x^(2)+a_(3)x^(3)andf(x)=sqrt(g(x)),f(x)` have its non-zero local minimum and maximum values at -3 and 3 respectively. If `a_(3) in ` the domain of the function `h(x)=sin^(-1)((1+x^(2))/(2x))`
The value of `a_(1)+a_(2)` is equal to

Options

A. 30
B. -30
C. 27
D. -27

Answer

To solve the problem step by step, we will analyze the given information and derive the required values systematically.

### Step 1: Understanding the Functions
We have \( g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 \) and \( f(x) = \sqrt{g(x)} \). The function \( f(x) \) has local minimum and maximum values at \( x = -3 \) and \( x = 3 \) respectively.

### Step 2: Finding Derivatives
Since \( f(x) \) has local extrema at \( x = -3 \) and \( x = 3 \), we know that the derivative \( f'(x) \) must be zero at these points. Using the chain rule, we find:
\[ f'(x) = \frac{1}{2\sqrt{g(x)}} g'(x) \]
Setting \( f'(-3) = 0 \) and \( f'(3) = 0 \) implies \( g'(-3) = 0 \) and \( g'(3) = 0 \).

### Step 3: Finding \( g'(x) \)
The derivative \( g'(x) \) is given by:
\[ g'(x) = a_1 + 2a_2 x + 3a_3 x^2 \]
Thus, we have:
1. \( g'(-3) = a_1 - 6a_2 + 27a_3 = 0 \) (1)
2. \( g'(3) = a_1 + 6a_2 + 27a_3 = 0 \) (2)

### Step 4: Solving the System of Equations
We now have a system of equations from (1) and (2):
1. \( a_1 - 6a_2 + 27a_3 = 0 \)
2. \( a_1 + 6a_2 + 27a_3 = 0 \)

Subtracting (1) from (2):
\[ (a_1 + 6a_2 + 27a_3) - (a_1 - 6a_2 + 27a_3) = 0 \implies 12a_2 = 0 \implies a_2 = 0 \]

### Step 5: Finding \( a_1 \)
Substituting \( a_2 = 0 \) back into either equation, we use (1):
\[ a_1 + 27a_3 = 0 \implies a_1 = -27a_3 \]

### Step 6: Finding \( a_3 \)
We are given that \( a_3 \) is in the domain of the function \( h(x) = \sin^{-1}\left(\frac{1 + x^2}{2x}\right) \). The domain of \( h(x) \) requires:
\[ -1 \leq \frac{1 + x^2}{2x} \leq 1 \]
This leads to the condition \( x \neq 0 \) and \( x^2 - 2x + 1 \geq 0 \), which simplifies to \( (x-1)^2 \geq 0 \). The critical point is \( x = 1 \), so \( a_3 \) can be any real number except zero.

Assuming \( a_3 = 1 \) (as a valid choice):
\[ a_1 = -27 \cdot 1 = -27 \]

### Step 7: Final Calculation
Now, we need to find \( a_1 + a_2 \):
\[ a_1 + a_2 = -27 + 0 = -27 \]

### Conclusion
The value of \( a_1 + a_2 \) is \( \boxed{-27} \).