If the derivative of the function
`f(x)={{:(ax^(2)+b,xlt-1),(bx^(2)+ax+4,xge-1):}`
is every where continuous, then what are the values of a and b?

Options

A. a = 2, b = 3
B. a = 3, b = 2
C. `a = -2, b = -3`
D. `a=-3,b=-2`

Answer

To solve the problem, we need to ensure that the derivative of the piecewise function \( f(x) \) is continuous everywhere, particularly at the point where the two pieces meet, which is at \( x = -1 \).

### Step 1: Define the function pieces
The function is defined as:
- For \( x < -1 \): \( f(x) = ax^2 + b \)
- For \( x \geq -1 \): \( f(x) = bx^2 + ax + 4 \)

### Step 2: Ensure continuity at \( x = -1 \)
For \( f(x) \) to be continuous at \( x = -1 \), the two pieces must equal each other at that point:
\[ f(-1) \text{ from the left} = f(-1) \text{ from the right} \]
Calculating \( f(-1) \) from the left:
\[ f(-1) = a(-1)^2 + b = a + b \]
Calculating \( f(-1) \) from the right:
\[ f(-1) = b(-1)^2 + a(-1) + 4 = b - a + 4 \]
Setting these equal gives:
\[ a + b = b - a + 4 \]

### Step 3: Solve for \( a \) and \( b \)
Rearranging the equation:
\[ a + b = b - a + 4 \implies a + a = 4 \implies 2a = 4 \implies a = 2 \]

### Step 4: Substitute \( a \) back to find \( b \)
Now substituting \( a = 2 \) back into the equation:
\[ 2 + b = b - 2 + 4 \]
This simplifies to:
\[ 2 + b = b + 2 \]
This equation holds for any \( b \), so we need to find the derivative continuity next.

### Step 5: Derivatives of both pieces
Now we find the derivatives:
- For \( x < -1 \):
\[ f'(x) = \frac{d}{dx}(ax^2 + b) = 2ax = 2(2)x = 4x \]
- For \( x \geq -1 \):
\[ f'(x) = \frac{d}{dx}(bx^2 + ax + 4) = 2bx + a = 2bx + 2 \]

### Step 6: Ensure derivative continuity at \( x = -1 \)
Setting the derivatives equal at \( x = -1 \):
\[ f'(-1) \text{ from the left} = f'(-1) \text{ from the right} \]
Calculating:
\[ f'(-1) \text{ from the left} = 4(-1) = -4 \]
\[ f'(-1) \text{ from the right} = 2b(-1) + 2 = -2b + 2 \]
Setting these equal gives:
\[ -4 = -2b + 2 \]
Rearranging:
\[ -4 - 2 = -2b \implies -6 = -2b \implies b = 3 \]

### Conclusion
The values of \( a \) and \( b \) are:
\[ a = 2, \quad b = 3 \]