If a differentiable function f(x) has a relative minimum at x=0, then the function g = f(x) + ax + b has a relative minimum at x =0 for

Options

A. all a and all b
B. all b if a =0
C. all b `gt` 0
D. all`a gt 0`

Answer

To solve the problem, we need to analyze the conditions under which the function \( g(x) = f(x) + ax + b \) has a relative minimum at \( x = 0 \), given that \( f(x) \) has a relative minimum at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding the conditions for \( f(x) \)**:
- Since \( f(x) \) has a relative minimum at \( x = 0 \), we know:
- \( f'(0) = 0 \) (the first derivative at this point is zero).
- \( f''(0) > 0 \) (the second derivative at this point is positive).

2. **Finding the first derivative of \( g(x) \)**:
- We differentiate \( g(x) \):
\[ g'(x) = f'(x) + a \]
- To find the condition for \( g(x) \) to have a relative minimum at \( x = 0 \), we need \( g'(0) = 0 \):
\[ g'(0) = f'(0) + a = 0 \]
- Substituting \( f'(0) = 0 \):
\[ 0 + a = 0 \implies a = 0 \]

3. **Finding the second derivative of \( g(x) \)**:
- Now we differentiate \( g'(x) \) to find \( g''(x) \):
\[ g''(x) = f''(x) \]
- To ensure that \( g(x) \) has a relative minimum at \( x = 0 \), we need \( g''(0) > 0 \):
\[ g''(0) = f''(0) > 0 \]
- This condition is already satisfied since \( f(x) \) has a relative minimum at \( x = 0 \).

4. **Conclusion**:
- For \( g(x) \) to have a relative minimum at \( x = 0 \), we have determined that:
- \( a = 0 \)
- \( b \) can be any real number (since it does not affect the derivatives).
- Therefore, the conditions are:
- \( a = 0 \)
- \( b \) is arbitrary.

### Final Answer:
For the function \( g(x) = f(x) + ax + b \) to have a relative minimum at \( x = 0 \), it is required that \( a = 0 \) and \( b \) can be any real number.