Find the values of `aa n db` so that the function `f(x)={x^2+3x+a ,ifxlt=1b x+2,ifx >1` is differentiable at each `x in Rdot`

Explanation

To find the values of \( a \) and \( b \) such that the function

\[ f(x) = \begin{cases} x^2 + 3x + a & \text{if } x \leq 1 \\ bx + 2 & \text{if } x > 1 \end{cases} \]

is differentiable at \( x = 1 \), we need to ensure that the function is continuous and that the derivatives from both sides at \( x = 1 \) are equal.

### Step 1: Ensure Continuity at \( x = 1 \)

For \( f(x) \) to be continuous at \( x = 1 \), we need:

\[ f(1^-) = f(1^+) \]

Calculating \( f(1^-) \):

\[ f(1^-) = 1^2 + 3(1) + a = 1 + 3 + a = 4 + a \]

Calculating \( f(1^+) \):

\[ f(1^+) = b(1) + 2 = b + 2 \]

Setting these equal for continuity:

\[ 4 + a = b + 2 \]

Rearranging gives us:

\[ b - a = 2 \quad \text{(Equation 1)} \]

### Step 2: Ensure Differentiability at \( x = 1 \)

For \( f(x) \) to be differentiable at \( x = 1 \), we need:

\[ f'(1^-) = f'(1^+) \]

Calculating \( f'(1^-) \):

The derivative of \( f(x) = x^2 + 3x + a \) is:

\[ f'(x) = 2x + 3 \]

Thus,

\[ f'(1^-) = 2(1) + 3 = 2 + 3 = 5 \]

Calculating \( f'(1^+) \):

The derivative of \( f(x) = bx + 2 \) is:

\[ f'(x) = b \]

Thus,

\[ f'(1^+) = b \]

Setting these equal for differentiability:

\[ 5 = b \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

From Equation 2, we have:

\[ b = 5 \]

Substituting \( b = 5 \) into Equation 1:

\[ 5 - a = 2 \]

Rearranging gives:

\[ a = 5 - 2 = 3 \]

### Final Values

Thus, the values of \( a \) and \( b \) are:

\[ a = 3, \quad b = 5 \]

### Summary

The values of \( a \) and \( b \) such that the function is differentiable at \( x = 1 \) are:

\[ \boxed{a = 3, b = 5} \]