Question:

Step 1: Apply continuity at $x = 1$ For $f(x)$ to be continuous at $x=1$, the left-hand limit must equal the right-hand limit: $$ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) $$ $$ a(1)^2 - b = 1 + b $$ $$ a - b = 1 + b $$ $$ a = 1 + 2b \quad (1) $$ Step 2: Apply differentiability at $x = 1$ For $f(x)$ to be differentiable at $x=1$, the left-hand derivative must equal the right-hand derivative. The derivative of the first piece is $ \frac{d}{dx}(ax^2 - b) = 2ax $. The derivative of the second piece is $ \frac{d}{dx}(x + b) = 1 $. Equating the derivatives at $x=1$: $$ 2a(1) = 1 $$ $$ 2a = 1 $$ $$ a = \frac{1}{2} \quad (2) $$ Step 3: Determine the value of $b$ Substitute the value of $a$ from equation (2) into equation (1): $$ \frac{1}{2} = 1 + 2b $$ $$ 2b = \frac{1}{2} - 1 $$ $$ 2b = -\frac{1}{2} $$ $$ b = -\frac{1}{4} $$ Step 4: Calculate $f(1)$ Using the definition of $f(x)$ for $x \ge 1$, we have $f(x) = x+b$. Substitute $x=1$ and the value of $b$: $$ f(1) = 1 + b = 1 + \left(-\frac{1}{4}\right) = 1 - \frac{1}{4} = \frac{3}{4} $$ Alternatively, using the definition of $f(x)$ for $x < 1$, we have $f(x) = ax^2 - b$. Substitute $x=1$ and the values of $a$ and $b$: $$ f(1) = a(1)^2 - b = \frac{1}{2}(1)^2 - \left(-\frac{1}{4}\right) = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4} $$ Both definitions yield $f(1) = \frac{3}{4}$.