Let `A =[a, oo)` denotes domain, then `f:[a, oo) to B, f(x) =2x ^(3)+6` will have an inverse for then smallest real values of a, if:

Options

A. `a=1, B=[5,oo)`
B. `a=2,B=[10,oo)`
C. `a = 0,B=[6,oo)`
D. `a=-1,B=[1,oo)`

Answer

To find the smallest real value of \( a \) such that the function \( f(x) = 2x^3 + 6 \) has an inverse, we need to ensure that \( f \) is a one-to-one function on the domain \( [a, \infty) \). A function is one-to-one if it is either strictly increasing or strictly decreasing.

### Step-by-step Solution:

1. **Determine the derivative of the function**:
\[ f'(x) = \frac{d}{dx}(2x^3 + 6) = 6x^2 \]

2. **Analyze the derivative**:
The derivative \( f'(x) = 6x^2 \) is non-negative for all \( x \) in the domain \( [a, \infty) \). It is zero only at \( x = 0 \) and positive for all \( x > 0 \). This means that \( f(x) \) is strictly increasing for \( x > 0 \).

3. **Determine the behavior at \( x = 0 \)**:
If we want \( f \) to be one-to-one starting from \( a \), we need to ensure that \( a \) is at least 0. If \( a < 0 \), then \( f \) would not be strictly increasing over the entire interval \( [a, \infty) \).

4. **Check the value of \( f(a) \)**:
\[ f(0) = 2(0)^3 + 6 = 6 \]
Therefore, when \( a = 0 \), the function starts at \( f(0) = 6 \).

5. **Determine the range of \( f \)**:
As \( x \) approaches infinity, \( f(x) \) approaches infinity. Thus, the range of \( f \) when \( a = 0 \) is \( [6, \infty) \).

6. **Determine the domain of the inverse function**:
The inverse function \( f^{-1}(x) \) will have a domain equal to the range of \( f \). Therefore, the domain of \( f^{-1}(x) \) is \( [6, \infty) \).

7. **Conclusion**:
The smallest real value of \( a \) such that \( f \) has an inverse is \( a = 0 \).

### Final Answer:
The smallest real value of \( a \) is \( 0 \) and the range \( B \) is \( [6, \infty) \).

---