If `y=a log x +bx^(2)+x` has its exteme values at x=-1 and x=2, then find a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the function \( y = a \log x + b x^2 + x \) has extreme values at \( x = -1 \) and \( x = 2 \).

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

\[ y = a \log x + b x^2 + x \]

The derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = \frac{a}{x} + 2bx + 1 \]

### Step 2: Set the derivative to zero at the extreme points
Since we know that the extreme values occur at \( x = -1 \) and \( x = 2 \), we will set the derivative equal to zero at these points.

#### For \( x = -1 \):
\[ \frac{dy}{dx} \bigg|_{x=-1} = \frac{a}{-1} + 2b(-1) + 1 = -a - 2b + 1 = 0 \]

This simplifies to:
\[ -a - 2b + 1 = 0 \quad \Rightarrow \quad a + 2b = 1 \quad \text{(Equation 1)} \]

#### For \( x = 2 \):
\[ \frac{dy}{dx} \bigg|_{x=2} = \frac{a}{2} + 2b(2) + 1 = \frac{a}{2} + 4b + 1 = 0 \]

This simplifies to:
\[ \frac{a}{2} + 4b + 1 = 0 \quad \Rightarrow \quad a + 8b = -2 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations
Now we have a system of two equations:

1. \( a + 2b = 1 \)
2. \( a + 8b = -2 \)

We can solve these equations simultaneously.

#### Subtract Equation 1 from Equation 2:
\[ (a + 8b) - (a + 2b) = -2 - 1 \]
This simplifies to:
\[ 6b = -3 \quad \Rightarrow \quad b = -\frac{1}{2} \]

### Step 4: Substitute \( b \) back to find \( a \)
Now that we have \( b \), we can substitute it back into Equation 1 to find \( a \):

\[ a + 2(-\frac{1}{2}) = 1 \]
This simplifies to:
\[ a - 1 = 1 \quad \Rightarrow \quad a = 2 \]

### Final Result
Thus, the values of \( a \) and \( b \) are:
\[ a = 2, \quad b = -\frac{1}{2} \]