Find the values of a and b that make the piecewise function f(x) continuous for all real numbers. f(x) = 1+x if x 5, ax+b if 5 < x < 12, x+2 if x 12

For continuity at $x=5$, we must have $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x) = f(5)$.

$$\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (1+x) = 1+5 = 6$$ $$\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} (ax+b) = a(5)+b = 5a+b$$ $$f(5) = 1+5 = 6$$

Equating the limits gives our first equation:

$$5a+b = 6 \quad \text{(Equation 1)}$$

For continuity at $x=12$, we must have $\lim_{x \to 12^-} f(x) = \lim_{x \to 12^+} f(x) = f(12)$.

$$\lim_{x \to 12^-} f(x) = \lim_{x \to 12^-} (ax+b) = a(12)+b = 12a+b$$ $$\lim_{x \to 12^+} f(x) = \lim_{x \to 12^+} (x+2) = 12+2 = 14$$ $$f(12) = 12+2 = 14$$

Equating the limits gives our second equation:

$$12a+b = 14 \quad \text{(Equation 2)}$$

We have the system:

1. $5a+b = 6$
2. $12a+b = 14$

Subtract Equation 1 from Equation 2:

$$(12a+b) - (5a+b) = 14 - 6$$ $$7a = 8$$ $$a = \frac{8}{7}$$

$$5\left(\frac{8}{7}\right) + b = 6$$ $$\frac{40}{7} + b = 6$$ $$b = 6 - \frac{40}{7}$$

To subtract, find a common denominator:

$$b = \frac{42}{7} - \frac{40}{7}$$ $$b = \frac{2}{7}$$

The values of $a$ and $b$ that make the function continuous are:

$$a = \boxed{\frac{8}{7}}$$ $$b = \boxed{\frac{2}{7}}$$