For what choices of a, b, c, if any, does the function
`f(x)={{:(ax^(2)+bx+x" , "0lexle1),(bx-c" , "1lexle2),(x" , "xgt2):}`
become differentiable at x = 1 and x = 2?

Answer

To determine the values of \( a \), \( b \), and \( c \) such that the function

\[ f(x) = \begin{cases} ax^2 + bx + 1 & \text{if } 0 \leq x \leq 1 \\ bx - c & \text{if } 1 < x \leq 2 \\ x & \text{if } x > 2 \end{cases} \]

is differentiable at \( x = 1 \) and \( x = 2 \), we need to ensure that the function is continuous and that the derivatives from the left and right at these points are equal.

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

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

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1) \]

Calculating these limits:

1. **Left limit** as \( x \to 1^- \):
\[ f(1) = a(1)^2 + b(1) + 1 = a + b + 1 \]

2. **Right limit** as \( x \to 1^+ \):
\[ f(1) = b(1) - c = b - c \]

Setting these equal for continuity:
\[ a + b + 1 = b - c \]

This simplifies to:
\[ a + c + 1 = 0 \quad \text{(Equation 1)} \]

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

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

\[ \lim_{x \to 1^-} f'(x) = \lim_{x \to 1^+} f'(x) \]

Calculating the derivatives:

1. **Left derivative**:
\[ f'(x) = 2ax + b \quad \Rightarrow \quad f'(1) = 2a(1) + b = 2a + b \]

2. **Right derivative**:
\[ f'(x) = b \quad \Rightarrow \quad f'(1) = b \]

Setting these equal for differentiability:
\[ 2a + b = b \]

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

### Step 3: Check Continuity at \( x = 2 \)

For \( f(x) \) to be continuous at \( x = 2 \), we need:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2) \]

Calculating these limits:

1. **Left limit** as \( x \to 2^- \):
\[ f(2) = b(2) - c = 2b - c \]

2. **Right limit** as \( x \to 2^+ \):
\[ f(2) = 2 \]

Setting these equal for continuity:
\[ 2b - c = 2 \quad \text{(Equation 3)} \]

### Step 4: Solve the System of Equations

Now we have the following equations to solve:

1. \( a + c + 1 = 0 \) (from continuity at \( x = 1 \))
2. \( a = 0 \) (from differentiability at \( x = 1 \))
3. \( 2b - c = 2 \) (from continuity at \( x = 2 \))

Substituting \( a = 0 \) into Equation 1:
\[ 0 + c + 1 = 0 \quad \Rightarrow \quad c = -1 \]

Now substituting \( c = -1 \) into Equation 3:
\[ 2b - (-1) = 2 \quad \Rightarrow \quad 2b + 1 = 2 \quad \Rightarrow \quad 2b = 1 \quad \Rightarrow \quad b = \frac{1}{2} \]

### Final Values

Thus, we have:
- \( a = 0 \)
- \( b = \frac{1}{2} \)
- \( c = -1 \)

### Summary of Values
The values of \( a \), \( b \), and \( c \) that make the function differentiable at \( x = 1 \) and \( x = 2 \) are:
- \( a = 0 \)
- \( b = \frac{1}{2} \)
- \( c = -1 \)