For what values of a and b, is the function `f(x)= {(x^(2) "," x le c),(ax + b"," x gt c):}` differentiable at x=c

Answer

To determine the values of \( a \) and \( b \) for which the function

\[ f(x) = \begin{cases} x^2 & \text{if } x \leq c \\ ax + b & \text{if } x > c \end{cases} \]

is differentiable at \( x = c \), we need to ensure that the function is both continuous and differentiable at that point.

### Step 1: Check for Continuity at \( x = c \)

For the function to be continuous at \( x = c \), the left-hand limit must equal the right-hand limit at that point.

- **Left-hand limit** as \( x \) approaches \( c \) from the left:
\[ \lim_{x \to c^-} f(x) = c^2 \]

- **Right-hand limit** as \( x \) approaches \( c \) from the right:
\[ \lim_{x \to c^+} f(x) = ac + b \]

Setting these two limits equal gives us the first equation:
\[ ac + b = c^2 \quad \text{(Equation 1)} \]

### Step 2: Check for Differentiability at \( x = c \)

For the function to be differentiable at \( x = c \), the left-hand derivative must equal the right-hand derivative at that point.

- **Left-hand derivative**:
\[ f'(x) = 2x \quad \text{(for } x \leq c\text{)} \]
Evaluating at \( x = c \):
\[ f'(c) = 2c \]

- **Right-hand derivative**:
\[ f'(x) = a \quad \text{(for } x > c\text{)} \]
Thus, we have:
\[ f'(c) = a \]

Setting these two derivatives equal gives us the second equation:
\[ a = 2c \quad \text{(Equation 2)} \]

### Step 3: Solve the Equations

Now we have two equations:
1. \( ac + b = c^2 \)
2. \( a = 2c \)

Substituting Equation 2 into Equation 1:
\[ (2c)c + b = c^2 \]
This simplifies to:
\[ 2c^2 + b = c^2 \]
Rearranging gives:
\[ b = c^2 - 2c^2 = -c^2 \]

### Final Values

Thus, we find:
\[ a = 2c \quad \text{and} \quad b = -c^2 \]

### Summary of Solution
The values of \( a \) and \( b \) for which the function is differentiable at \( x = c \) are:
\[ a = 2c, \quad b = -c^2 \]