1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} x - a, & x < 0 \\ a x^2 + b x + 1, & 0 \leq x \leq 1 \\ x^2 + 2x, & x > 1 \end{cases}$$ and we want to find the value of $b - a$ such that $f(x)$ is continuous for all real $x$. 2. **Recall the continuity condition:** A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are all equal. 3. **Check continuity at $x=0$:** - Left limit: $\lim_{x \to 0^-} f(x) = 0 - a = -a$ - Right limit: $\lim_{x \to 0^+} f(x) = a \cdot 0^2 + b \cdot 0 + 1 = 1$ - Function value at 0: $f(0) = 1$ Set left and right limits equal for continuity: $$-a = 1 \implies a = -1$$ 4. **Check continuity at $x=1$:** - Left limit: $\lim_{x \to 1^-} f(x) = a \cdot 1^2 + b \cdot 1 + 1 = a + b + 1$ - Right limit: $\lim_{x \to 1^+} f(x) = 1^2 + 2 \cdot 1 = 3$ - Function value at 1: $f(1) = a + b + 1$ Set left and right limits equal for continuity: $$a + b + 1 = 3$$ Substitute $a = -1$: $$-1 + b + 1 = 3 \implies b = 3$$ 5. **Calculate $b - a$:** $$b - a = 3 - (-1) = 3 + 1 = 4$$ **Final answer:** $b - a = 4$