Exercise 5.1 · Q30: Find $a$⁠, $b$⁠: 5, then $ax+b$⁠, then 21 is continuous

Basics you need

Solution

1. The domain is all of $\R$⁠, split at 2 and 10

   Every real number satisfies exactly one of $x \le 2$⁠, $2 < x < 10$⁠, $x \ge 10$⁠, so $f$ is defined at every real number. The points to examine are

   • points $c < 2$⁠, points with $2 < c < 10$⁠, points $c > 10$ (inside a piece);
   • the joints $c = 2$ and $c = 10$⁠.
   "Continuous function" means continuous at every point of the domain (Definition 2). Splitting the line this way covers every real number exactly once, so nothing is missed.
   Read the inequalities carefully: the three pieces meet at 2 and 10, and each of those points belongs to exactly one piece (Basic 3). Basics: piecewise functions

2. Inside each piece, $f$ is continuous for every $a$⁠, $b$

   • If $c < 2$⁠: every $x$ close enough to $c$ (closer than $2 - c$⁠) also satisfies $x < 2$⁠, so near $c$ the function is the constant 5. Hence $\lim_{x\to c} f(x) = 5 = f(c)$⁠.
   • If $2 < c < 10$⁠: every $x$ close enough to $c$ also lies strictly between 2 and 10, so near $c$⁠, $f(x) = ax + b$⁠, a polynomial. Hence $\lim_{x\to c} f(x) = ac + b = f(c)$⁠.
   • If $c > 10$⁠: near $c$ the function is the constant 21, so $\lim_{x\to c} f(x) = 21 = f(c)$⁠.

   So $f$ is continuous at every point other than 2 and 10, whatever the values of $a$ and $b$⁠.

   A limit at $c$ depends only on the values of $f$ near $c$⁠. For a point strictly inside a piece we can stay so close to $c$ that we never leave that piece, and there $f$ is a constant or a straight line, both continuous. That is why only the joints can cause trouble.
   Constants: Example 5; polynomials: Example 14. Basics: what a limit is

3. The joint $x = 2$⁠: need $2a + b = 5$

   Value: $2 \le 2$⁠, so the first line applies and $f(2) = 5$⁠.

   Left-hand limit (only $x < 2$⁠, first line):

   $$\lim_{x\to2^-} f(x) = \lim_{x\to2^-} 5 = 5.$$

   Right-hand limit ($x$ just above 2 lies in $2 < x < 10$⁠, middle line):

   $$\lim_{x\to2^+} f(x) = \lim_{x\to2^+} (ax + b) = 2a + b.$$

   Continuity at 2 needs $5 = 2a + b = 5$⁠, that is

   $$2a + b = 5. \qquad (1)$$
   The middle line is not used at $x = 2$ itself (the inequality $2 < x$ is strict), but it is what $f$ follows as $x$ comes down to 2 from the right. So it supplies the right-hand limit, and that has to match the value 5.
   Definition 1 at $c = 2$⁠. Basics: left-hand and right-hand limits

4. The joint $x = 10$⁠: need $10a + b = 21$

   Value: $10 \ge 10$⁠, so the third line applies and $f(10) = 21$⁠.

   Left-hand limit ($x$ just below 10 lies in $2 < x < 10$⁠, middle line):

   $$\lim_{x\to10^-} f(x) = \lim_{x\to10^-} (ax + b) = 10a + b.$$

   Right-hand limit (only $x > 10$⁠, third line):

   $$\lim_{x\to10^+} f(x) = \lim_{x\to10^+} 21 = 21.$$

   Continuity at 10 needs $10a + b = 21 = 21$⁠, that is

   $$10a + b = 21. \qquad (2)$$
   This time the middle line approaches the joint from the left. It has to arrive at the height 21 where the last piece is sitting.
   Definition 1 at $c = 10$⁠. Lesson: continuity at a point

5. Solve the two equations

   Subtract (1) from (2): the $b$'s cancel.

   $$\begin{aligned} (10a + b) - (2a + b) &= 21 - 5\\ 8a &= 16\\ a &= 2. \end{aligned}$$

   Put $a = 2$ into (1): $4 + b = 5$⁠, so $b = 1$⁠.

   The two equations have exactly one solution, so $a = 2$⁠, $b = 1$ is the only choice. With any other pair, (1) or (2) fails and $f$ jumps at 2 or at 10.

   Both conditions are needed at once: $f$ must be continuous at 2 and at 10. Two conditions and two unknowns give one answer here because the lines in the $(a, b)$-plane, $2a + b = 5$ and $10a + b = 21$⁠, have different slopes and so cross at a single point.
   Elimination: subtract to remove one unknown (Basic 5).

6. Confirm with $a = 2$⁠, $b = 1$

   The middle piece is now $2x + 1$⁠:

   Joint	Left-hand limit	Right-hand limit	Value
   $x = 2$	5	$2(2) + 1 = 5$	$f(2) = 5$
   $x = 10$	$2(10) + 1 = 21$	21	$f(10) = 21$

   All agree, and Step 2 covers every other point, so $f$ is continuous at every real number. ✓

   The picture makes the conditions concrete: the blue segment is the line through $(2, 5)$ and $(10, 21)$⁠. Its slope is $\frac{21 - 5}{10 - 2} = \frac{16}{8} = 2$⁠, the value of $a$ we found.

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