If the function `y = ae^(x) + bx^(2)+3x` is maximum at x = 0 and minimum at x = - 3, then find the values of a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) for the function \( y = ae^x + bx^2 + 3x \) given that it has a maximum at \( x = 0 \) and a minimum at \( x = -3 \).

### Step 1: Find the first derivative of the function
The first step is to differentiate the function \( y \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(ae^x) + \frac{d}{dx}(bx^2) + \frac{d}{dx}(3x) \]

Using the rules of differentiation, we find:

\[ \frac{dy}{dx} = ae^x + 2bx + 3 \]

### Step 2: Set the first derivative to zero at critical points
Since the function has a maximum at \( x = 0 \) and a minimum at \( x = -3 \), we set the first derivative equal to zero at these points.

1. At \( x = 0 \):
\[ \frac{dy}{dx}\bigg|_{x=0} = ae^0 + 2b(0) + 3 = a + 3 = 0 \]
This gives us our first equation:
\[ a + 3 = 0 \quad \Rightarrow \quad a = -3 \]

2. At \( x = -3 \):
\[ \frac{dy}{dx}\bigg|_{x=-3} = ae^{-3} + 2b(-3) + 3 = ae^{-3} - 6b + 3 = 0 \]
Substituting \( a = -3 \) into this equation:
\[ -3e^{-3} - 6b + 3 = 0 \]
Rearranging gives us:
\[ -6b = 3e^{-3} - 3 \]
\[ b = \frac{3 - 3e^{-3}}{6} \]
\[ b = \frac{1 - e^{-3}}{2} \]

### Step 3: Final values of \( a \) and \( b \)
Thus, we have found the values:
- \( a = -3 \)
- \( b = \frac{1 - e^{-3}}{2} \)

### Summary of the solution
The values of \( a \) and \( b \) are:
\[ a = -3, \quad b = \frac{1 - e^{-3}}{2} \]