Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q27: Find $k$ ⁠: $kx^2$ for $x\le2$ ⁠, 3 for $x>2$

Exercise 5.1Continuity at a point · Section 5.2

The question

Find the values of $k$ so that the function $f$ is continuous at the indicated point in Exercises 26 to 29.

27. $f(x) = \begin{cases} kx^2, & \text{if } x \le 2 \\ 3, & \text{if } x > 2 \end{cases} \qquad \text{at } x = 2.$

Exam answer

\begin{aligned} f(2) &= k(2)^2 = 4k \\ \text{LHL} &= \lim_{h\to0}f(2 - h) \\ &= \lim_{h\to0}k(2 - h)^2 = 4k \\ \text{RHL} &= \lim_{h\to0}f(2 + h) \\ &= \lim_{h\to0}3 = 3 \end{aligned}

For continuity at $x = 2$ ⁠,

\text{LHL} = \text{RHL} = f(2)

\Rightarrow\ 4k = 3\ \Rightarrow\ k = \frac34

$\therefore\ k = \dfrac34$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The two pieces must meet at $x = 2$ ⁠. Coming from the left, $f$ follows $kx^2$ ⁠, which heads to $k \cdot 2^2 = 4k$ (and this is also the value $f(2)$ ⁠, because $x = 2$ belongs to the first piece). Coming from the right, $f$ is the constant 3. The pieces join exactly when $4k = 3$ ⁠, so $k = \frac{3}{4}$ ⁠.

Given, to find, and the plan

Given: $f(x) = kx^2$ for $x \le 2$ and $f(x) = 3$ for $x > 2$ ⁠, with $k$ an unknown constant.
To find: Every value of $k$ for which $f$ is continuous at $x = 2$ ⁠.
Plan: (1) Find $f(2)$ ⁠, using the piece that contains 2. (2) Left-hand limit from $kx^2$ ⁠. (3) Right-hand limit from the constant 3. (4) Set all three equal and solve for $k$ ⁠. (5) Confirm with the value found.

Basics you need

1Continuity at a point

$f$ is continuous at a point $c$ of its domain when the left-hand limit, the right-hand limit and $f(c)$ all exist and are equal (Definition 1, Section 5.2). Lesson: continuity at a point

2Which piece gives $f(2)$ ⁠?

The sign " $\le$ " in " $x \le 2$ " includes 2 itself, so $f(2)$ comes from the first line. The second line, " $x > 2$ ", never applies at 2. Basics: piecewise functions

3One-sided limits

The left-hand limit at 2 only uses $x < 2$ (first line); the right-hand limit only uses $x > 2$ (second line). Basics: left-hand and right-hand limits

4Limits of polynomials and constants

A polynomial is continuous (Example 14), so its limit at any point is its value there. A constant's limit is the constant (Example 5). Basics: what a limit is

Solution

The value at the point: $f(2) = 4k$

Every real $x$ satisfies exactly one of $x \le 2$ or $x > 2$ ⁠, so $f$ is defined everywhere; in particular 2 is in the domain. Since $2 \le 2$ ⁠, the first line applies:
$f(2) = k \cdot 2^2 = 4k.$
Continuity compares the limit with the actual value, so we need the value first, from the correct line. Using "3" here would be the most common slip: 3 is the rule only for $x$ strictly greater than 2.

" $\le$ " means "less than or equal to"; " $>$ " means "strictly greater than" (Basic 2).
Left-hand limit: follow $kx^2$

For $x < 2$ ⁠, $f(x) = kx^2$ ⁠. Put $x = 2 - h$ with $h > 0$ ⁠, $h \to 0$ ⁠:
$\begin{aligned} \lim_{x\to2^-} f(x) &= \lim_{h\to0^+} k(2-h)^2\\ &= \lim_{h\to0^+} k\,(4 - 4h + h^2)\\ &= 4k. \end{aligned}$
To the left of 2 the function is the polynomial $kx^2$ ⁠, and a polynomial's limit is found by substituting (Example 14). Writing $x = 2 - h$ just makes the "approach from the left" visible: the terms $-4h$ and $h^2$ shrink to 0.

$(2-h)^2 = 4 - 4h + h^2$ ⁠. Basics: left-hand and right-hand limits
Right-hand limit: the constant 3

For $x > 2$ ⁠, $f(x) = 3$ ⁠, whatever $x$ is. So
$\lim_{x\to2^+} f(x) = \lim_{h\to0^+} 3 = 3.$
To the right of 2 the graph is the horizontal line $y = 3$ ⁠. A constant does not change as $x$ moves, so its limit is the constant itself (Example 5).

Limit of a constant function: $\lim_{x\to c} k = k$ ⁠. Basics: what a limit is
Set them equal: $4k = 3$

Continuity at 2 needs
$\lim_{x\to2^-} f(x) = \lim_{x\to2^+} f(x) = f(2),$
that is, $4k = 3 = 4k$ ⁠. The first and last are the same expression, so the only real condition is
$4k = 3 \quad\Longleftrightarrow\quad k = \frac{3}{4}.$
For any other $k$ ⁠, the left side approaches $4k \ne 3$ while the right side approaches 3: the graph jumps at $x = 2$ ⁠. So $k = \frac{3}{4}$ is the only value.

Because $x = 2$ belongs to the left piece, the value $f(2)$ automatically matches the left-hand limit. All the work is in making the right-hand limit agree too.

Definition 1: left-hand limit = right-hand limit = value. Lesson: continuity at a point
Confirm with $k = \frac{3}{4}$

Now $f(x) = \frac{3}{4}x^2$ for $x \le 2$ and $f(x) = 3$ for $x > 2$ ⁠:
$\begin{aligned} \lim_{x\to2^-} f(x) &= \tfrac{3}{4}\cdot 4 = 3,\\ \lim_{x\to2^+} f(x) &= 3,\\ f(2) &= \tfrac{3}{4}\cdot 4 = 3. \end{aligned}$
All three are 3, so $f$ is continuous at $x = 2$ ⁠. ✓

With $k = \frac{3}{4}$ the parabola $y = \frac{3}{4}x^2$ (blue) ends at the filled dot $(2, 3)$ ⁠, exactly where the line $y = 3$ (gold) begins. The graph can be drawn through $x = 2$ without lifting the pen.

Substituting the answer back is a cheap way to catch an arithmetic slip, and the graph shows what the equation $4k = 3$ means: the height of the parabola at $x = 2$ has to be the height of the line.

Result

$f$ is continuous at $x = 2$ if and only if $k = \frac{3}{4}$ ; then both one-sided limits and $f(2)$ equal 3. (Textbook answer: $k = \frac{3}{4}$ ⁠.)

Check it with numbers

With $k = \frac{3}{4} = 0.75$ ⁠:

\begin{aligned} f(1.99) &= 0.75 \times 3.9601 = 2.970075,\\ f(1.999) &= 0.75 \times 3.996001 = 2.99700075, \end{aligned}

while $f(2.001) = 3$ and $f(2) = 0.75 \times 4 = 3$ ⁠. The left values climb to 3 and the right values are 3. ✓ With $k = 1$ instead, $f(1.999) \approx 3.996$ ⁠, near 4, so the graph would jump from 4 down to 3.

Common mistakes

Taking $f(2) = 3$ ⁠. The line "3, if $x > 2$ " does not include 2. $f(2) = 4k$ ⁠.
Writing $k \cdot 2 = 3$ ⁠. The rule is $kx^2$ ⁠, so at 2 it is $k\cdot 4$ ⁠, not $2k$ ⁠.
Stopping at " $4k = 3$ ". Solve it: $k = \frac{3}{4}$ ⁠, and say it is the only value.

What to take away

For a two-piece rule, continuity at the joint means: the two pieces give the same height there.
Read the inequality signs to see which piece owns the joint; that piece also gives $f(c)$ ⁠.
One unknown, one condition: solve and state that the value is unique.

Every fact used, and where it comes from

Fact	Used in	Source
Continuity at $c$ ⁠: left-hand limit, right-hand limit and $f(c)$ equal	Steps 4, 5	Definition 1, Section 5.2 (Lesson: continuity at a point)
Polynomials are continuous, so their limit is their value	Step 2	Example 14, Section 5.2
A constant function's limit is the constant	Step 3	Example 5, Section 5.2
$(2-h)^2 = 4 - 4h + h^2$	Step 2	Algebra: $(a-b)^2 = a^2 - 2ab + b^2$

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Exercise 5.1 · Q27: Find kkk⁠: kx2kx^2kx2 for x≤2x\le2x≤2⁠, 3 for x>2x>2x>2

Basics you need

Solution

The value at the point: f(2)=4kf(2) = 4kf(2)=4k

Left-hand limit: follow kx2kx^2kx2

Right-hand limit: the constant 3

Set them equal: 4k=34k = 34k=3

Confirm with k=34k = \frac{3}{4}k=43​

Every fact used, and where it comes from

Exercise 5.1 · Q27: Find $k$ ⁠: $kx^2$ for $x\le2$ ⁠, 3 for $x>2$

The value at the point: $f(2) = 4k$

Left-hand limit: follow $kx^2$

Set them equal: $4k = 3$

Confirm with $k = \frac{3}{4}$