Exercise 5.1 · Q17: Joining $ax+1$ and $bx+3$ at $x=3$

Basics you need

Solution

1. Check that $3$ is in the domain and find $f(3)$

   Every real number satisfies exactly one of "$x\le3$" and "$x\gt3$", so $f$ is defined on all of $\R$⁠; in particular $3$ is in the domain. Since $3\le3$ is true, the first rule applies:

   $$f(3)=a(3)+1=3a+1.$$
   Continuity is only discussed at points of the domain, so we confirm that first. The sign "$\le$" in the first line is what puts the point $3$ itself on the left piece. The rule $bx+3$ is only for $x$ strictly greater than $3$⁠, so it never gives the value at $3$⁠.
   Which line of a piecewise definition to use is decided by the condition, not by convenience. Basics: piecewise functions

2. The left-hand limit at $3$

   Approach $3$ from the left: put $x=3-h$ with $h\gt0$ and let $h\to0$⁠. Such $x$ are less than $3$⁠, so the first rule applies:

   $$f(3-h)=a(3-h)+1=3a+1-ah.$$

   As $h\to0$⁠, the term $ah\to0$⁠, so

   $$\lim_{x\to3^-}f(x)=\lim_{h\to0}\,(3a+1-ah)=3a+1.$$
   The left-hand limit asks what height the graph is heading for as we walk towards $x=3$ from the left. On that side the graph is the line $y=ax+1$⁠, and a line heads straight to its own value at $3$⁠.
   $a$ is a fixed constant, so $ah$ shrinks to $0$ with $h$ (Basic 4). Basics: left-hand and right-hand limits

3. The right-hand limit at $3$

   Approach $3$ from the right: put $x=3+h$ with $h\gt0$⁠, $h\to0$⁠. Now $x\gt3$⁠, so the second rule applies:

   $$f(3+h)=b(3+h)+3=3b+3+bh.$$

   As $h\to0$⁠, $bh\to0$⁠, so

   $$\lim_{x\to3^+}f(x)=\lim_{h\to0}\,(3b+3+bh)=3b+3.$$
   The right-hand limit uses the formula $bx+3$ even though $f(3)$ does not. A limit only looks at points near $3$⁠, never at $3$ itself, and every point just to the right of $3$ is governed by the second rule.
   The value of a limit does not depend on the value of the function at the point. Basics: left-hand and right-hand limits

4. Demand that the three numbers agree

   By Definition 1, $f$ is continuous at $3$ exactly when

   $$\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)=f(3),$$

   that is, by Steps 1 to 3,

   $$3a+1=3b+3=3a+1.$$

   The first and last numbers are the same expression, so they are always equal. The only condition is the middle equality:

   $$\begin{aligned} 3a+1&=3b+3\\ 3a-3b&=2\\ a-b&=\tfrac{2}{3}\\ a&=b+\tfrac{2}{3}. \end{aligned}$$
   We subtract $3b$ and $1$ from both sides to collect the unknowns on the left, then divide by $3$⁠. Every step is reversible (we only added, subtracted or divided by a non-zero number), so the last line says exactly the same thing as the first.
   Continuity at a junction is "the left piece, the right piece and the value meet at one height". Lesson: continuity at a point

5. The relation is enough, not just necessary

   Step 4 shows: if $f$ is continuous at $3$⁠, then $a=b+\frac23$⁠. Conversely, suppose $a=b+\frac23$⁠. Then

   $$3a+1=3b+2+1=3b+3,$$

   so the left-hand limit, the right-hand limit and $f(3)$ all equal $3b+3$⁠, and $f$ is continuous at $3$⁠. So $f$ is continuous at $x=3$ if and only if $a=b+\frac23$.

   The relation leaves one free choice. For example $b=0,\ a=\frac23$⁠; or $b=1,\ a=\frac53$⁠; or $b=-\frac23,\ a=0$⁠. Each pair gives a different function, and each is continuous at $3$⁠.

   "Find the relationship" means a condition that works in both directions: every pair satisfying it must give continuity, and every pair giving continuity must satisfy it. One equation in two unknowns cannot fix both, which is why the answer is a relation and not a pair of numbers.
   With the relation in force, $f$ is in fact continuous at every real number: at any $c\lt3$ it equals the polynomial $ax+1$ on the open interval $x\lt3$ around $c$⁠, and at any $c\gt3$ it equals $bx+3$ on $x\gt3$ (polynomials are continuous, Example 14). Lesson: algebra of continuous functions

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