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_(0)` is

Options

A. equal to 50
B. greater than 54
C. less than 54
D. less than 50

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the Functions
We have two functions:
- \( g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 \)
- \( f(x) = \sqrt{g(x)} \)

We know that \( f(x) \) has local minimum and maximum values at \( x = -3 \) and \( x = 3 \) respectively.

### Step 2: Find the Derivative of \( f(x) \)
To find the critical points (local minima and maxima), we need to find the derivative of \( f(x) \):
\[ f'(x) = \frac{1}{2\sqrt{g(x)}} g'(x) \]
Setting \( f'(-3) = 0 \) and \( f'(3) = 0 \) implies \( g'(-3) = 0 \) and \( g'(3) = 0 \) since \( \sqrt{g(x)} \) cannot be zero.

### Step 3: Find the Derivative of \( g(x) \)
The derivative of \( g(x) \) is:
\[ g'(x) = a_1 + 2a_2 x + 3a_3 x^2 \]
Setting \( g'(-3) = 0 \) and \( g'(3) = 0 \):
1. \( g'(-3) = a_1 - 6a_2 + 27a_3 = 0 \) (Equation 1)
2. \( g'(3) = a_1 + 6a_2 + 27a_3 = 0 \) (Equation 2)

### Step 4: Solve the System of Equations
Subtract Equation 1 from Equation 2:
\[ (a_1 + 6a_2 + 27a_3) - (a_1 - 6a_2 + 27a_3) = 0 \]
This simplifies to:
\[ 12a_2 = 0 \implies a_2 = 0 \]

### Step 5: Substitute \( a_2 \) Back
Now substitute \( a_2 = 0 \) into either Equation 1 or Equation 2:
Using Equation 1:
\[ a_1 + 27a_3 = 0 \implies a_1 = -27a_3 \]

### Step 6: Determine the Value of \( a_3 \)
The problem states 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 can be analyzed for \( x > 0 \) and \( x < 0 \). For \( x > 0 \), we find that the expression is always positive and bounded.

### Step 7: Analyze the Domain
For \( x > 0 \):
\[ \frac{1+x^2}{2x} \geq -1 \implies 1 + x^2 \geq -2x \implies x^2 + 2x + 1 \geq 0 \implies (x+1)^2 \geq 0 \]
This is always true.

For \( x < 0 \):
\[ \frac{1+x^2}{2x} \leq 1 \implies 1 + x^2 \leq 2x \implies x^2 - 2x + 1 \leq 0 \implies (x-1)^2 \leq 0 \]
This implies \( x = 1 \) is the only solution, which is not in the domain for \( x < 0 \).

### Step 8: Conclusion on \( a_3 \)
Since \( a_3 \) must be a real number, we can take \( a_3 = 1 \) (as suggested in the transcript).

### Step 9: Find \( a_1 \)
Substituting \( a_3 = 1 \) back into the equation for \( a_1 \):
\[ a_1 = -27 \cdot 1 = -27 \]

### Step 10: Write \( g(x) \)
Now we have:
\[ g(x) = a_0 - 27x + x^3 \]

### Step 11: Find \( a_0 \)
To ensure \( g(-3) \) and \( g(3) \) are non-negative (since \( f(x) \) must be real):
1. \( g(-3) = a_0 + 81 - 27(-3) = a_0 + 81 + 81 = a_0 + 162 \geq 0 \)
2. \( g(3) = a_0 - 81 - 27(3) = a_0 - 81 - 81 = a_0 - 162 \geq 0 \)

From \( g(-3) \):
\[ a_0 + 162 \geq 0 \implies a_0 \geq -162 \]

From \( g(3) \):
\[ a_0 - 162 \geq 0 \implies a_0 \geq 162 \]

Thus, the value of \( a_0 \) must be at least \( 162 \).

### Final Answer
The value of \( a_0 \) is:
\[ \boxed{162} \]