Determine the values of a and b so that the piecewise function f is continuous, where f(x) = x + 7a for x at most -2, f(x) = 7ax + b for -2 < x < 1, and f(x) = x - 7b for x at least 1.

Step-by-step solution

1. 1

   Identify where continuity can actually fail. Each of the three branches is a polynomial, so it is continuous wherever it applies. Trouble can only arise at the two breakpoints $x=-2$ and $x=1$, where the formula switches. Continuity at a breakpoint means the left-hand limit, the right-hand limit and the function value all coincide — which here reduces to making the two neighbouring formulas agree at that point. Two breakpoints give two equations, matching the two unknowns.

2. 2

   Write the matching condition at $x=-2$. The first branch owns $x=-2$, so $f(-2)=-2+7a$. The middle branch supplies the right-hand limit:

   $\lim_{x\to -2^{+}}f(x)=7a(-2)+b=-14a+b$

   Setting them equal:

   $-2+7a=-14a+b\ \Longrightarrow\ 21a-b=2\ \Longrightarrow\ b=21a-2$

3. 3

   Write the matching condition at $x=1$. The middle branch supplies the left-hand limit and the third branch owns $x=1$:

   $\lim_{x\to 1^{-}}f(x)=7a+b,\qquad f(1)=1-7b$

   Setting them equal:

   $7a+b=1-7b\ \Longrightarrow\ 7a+8b=1$

4. 4

   Solve the two-equation system by substitution. Put $b=21a-2$ into $7a+8b=1$:

   $7a+8(21a-2)=1\ \Longrightarrow\ 7a+168a-16=1\ \Longrightarrow\ 175a=17\ \Longrightarrow\ a=\frac{17}{175}$

5. 5

   Back-substitute for $b$ and reduce. Using $b=21a-2$ with $21\cdot 17=357$:

   $b=\frac{357}{175}-2=\frac{357}{175}-\frac{350}{175}=\frac{7}{175}=\frac{1}{25}$

6. 6

   Check both junctions in exact arithmetic. With $a=\tfrac{17}{175}$ and $b=\tfrac{1}{25}$: at $x=-2$ the left branch gives $-2+7a=-\tfrac{33}{25}$ and the middle gives $-14a+b=-\tfrac{33}{25}$; at $x=1$ the middle gives $7a+b=\tfrac{18}{25}$ and the right branch gives $1-7b=\tfrac{18}{25}$. Both pairs match exactly as fractions, so $f$ is continuous everywhere.