Exercise 5.1 · Q29: Find $k$⁠: $kx+1$ for $x\le5$⁠, $3x-5$ for $x>5$

Basics you need

Solution

1. The value at the point: $f(5) = 5k + 1$

   Every real $x$ satisfies exactly one of $x \le 5$⁠, $x > 5$⁠, so $f$ is defined everywhere and 5 is in the domain. Since $5 \le 5$⁠, the first line applies:

   $$f(5) = k\cdot 5 + 1 = 5k + 1.$$
   The limits will be compared with this number, so it must come from the correct line.
   "$\le$" includes equality (Basic 2).

2. Left-hand limit: follow $kx + 1$

   For $x < 5$⁠, $f(x) = kx + 1$⁠. With $x = 5 - h$⁠, $h > 0$⁠, $h \to 0$⁠:

   $$\begin{aligned} \lim_{x\to5^-} f(x) &= \lim_{h\to0^+} \big(k(5 - h) + 1\big)\\ &= \lim_{h\to0^+} (5k + 1 - kh)\\ &= 5k + 1. \end{aligned}$$
   The term $-kh$ is a fixed number times something that shrinks to 0, so it disappears. This is just the continuity of the polynomial $kx + 1$ at 5.
   Polynomial: substitute to get the limit. Basics: what a limit is

3. Right-hand limit: follow $3x - 5$

   For $x > 5$⁠, $f(x) = 3x - 5$⁠. With $x = 5 + h$⁠, $h > 0$⁠, $h \to 0$⁠:

   $$\begin{aligned} \lim_{x\to5^+} f(x) &= \lim_{h\to0^+} \big(3(5 + h) - 5\big)\\ &= \lim_{h\to0^+} (10 + 3h)\\ &= 10. \end{aligned}$$
   The right-hand limit does not involve $k$ at all: it is fixed by the second line. So the value of $k$ must be chosen to bring the left line up (or down) to height 10.
   $3(5+h) - 5 = 15 + 3h - 5 = 10 + 3h$⁠. Basics: left-hand and right-hand limits

4. Set them equal: $5k + 1 = 10$

   Continuity at 5 needs

   $$\lim_{x\to5^-} f(x) = \lim_{x\to5^+} f(x) = f(5),$$

   that is, $5k + 1 = 10 = 5k + 1$⁠. The only condition is

   $$5k + 1 = 10 \;\Longleftrightarrow\; 5k = 9 \;\Longleftrightarrow\; k = \frac{9}{5}.$$

   For any other $k$ the left line arrives at $5k + 1 \ne 10$ and the graph jumps. So $k = \frac{9}{5}$ is the only value.

   The value $f(5)$ and the left-hand limit are the same expression because 5 belongs to the left piece; the equation makes the right-hand limit agree with them.
   Definition 1: left-hand limit = right-hand limit = value. Lesson: continuity at a point

5. Confirm with $k = \frac{9}{5}$

   Now the first piece is $\frac{9}{5}x + 1$⁠, and at $x = 5$ it gives $9 + 1 = 10$⁠:

   $$\begin{aligned} \lim_{x\to5^-} f(x) &= \tfrac{9}{5}\cdot 5 + 1 = 10,\\ \lim_{x\to5^+} f(x) &= 3\cdot 5 - 5 = 10,\\ f(5) &= 10. \end{aligned}$$

   All three agree, so $f$ is continuous at $x = 5$⁠. ✓

   Continuity only needs the heights to match. The slopes $\frac{9}{5}$ and 3 are different, which makes a corner at $x = 5$⁠; a corner is allowed for continuity. (Corners matter for differentiability, Section 5.3.)

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