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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q6: Discontinuities: 2x+32x+3 for x≤2,x\le2, 2x−32x-3 for x>2x>2

Exercise 5.1Points of discontinuity · Section 5.2
The question

Find all points of discontinuity of f,f, where ff is defined by f(x)={2x+3,if x≤22x−3,if x>2f(x) = \begin{cases} 2x + 3, & \text{if } x \le 2 \\ 2x - 3, & \text{if } x > 2 \end{cases}

Exam answer

For x<2x < 2⁠, f(x)=2x+3f(x) = 2x + 3⁠, and for x>2x > 2⁠, f(x)=2x−3f(x) = 2x - 3⁠. Both are polynomials, so ff is continuous at every x≠2x \ne 2⁠.

At x=2x = 2⁠:

LHL=lim⁡h→0f(2−h)=lim⁡h→0(2(2−h)+3)=7RHL=lim⁡h→0f(2+h)=lim⁡h→0(2(2+h)−3)=1\begin{aligned} \text{LHL} &= \lim_{h\to0} f(2 - h) \\ &= \lim_{h\to0}\bigl(2(2 - h) + 3\bigr) = 7 \\ \text{RHL} &= \lim_{h\to0} f(2 + h) \\ &= \lim_{h\to0}\bigl(2(2 + h) - 3\bigr) = 1 \end{aligned}

and f(2)=2(2)+3=7f(2) = 2(2) + 3 = 7⁠. Since LHL≠RHL\text{LHL} \ne \text{RHL}⁠, ff is not continuous at x=2x = 2⁠.

Hence x=2x = 2 is the only point of discontinuity of ff⁠.

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

The one idea this problem is built on

Split the real line at the point where the formula changes. Inside each piece ff is a polynomial, and polynomials are continuous, so there is nothing to find there. The only candidate is the junction x=2,x = 2, which must be tested with one-sided limits. There the two lines y=2x+3y = 2x + 3 and y=2x−3y = 2x - 3 are at different heights (7 and 1), so the graph jumps.

Given, to find, and the plan
Given
f(x)=2x+3f(x) = 2x + 3 for x≤2x \le 2 and f(x)=2x−3f(x) = 2x - 3 for x>2.x > 2.
To find
Every point of the domain at which ff is not continuous.
Plan
(1) Domain. (2) Points c<2.c < 2. (3) Points c>2.c > 2. (4) The junction c=2c = 2⁠: LHL, RHL, f(2).f(2). (5) Collect the answer.

Basics you need

1Continuity at a point; point of discontinuity

ff is continuous at cc (in its domain) when the left-hand limit, the right-hand limit and f(c)f(c) are all equal. If not, cc is a point of discontinuity (Definition 1, Section 5.2). Lesson: continuity at a point

2Polynomials are continuous

Every polynomial is continuous at every real number (Example 14). Both 2x+32x + 3 and 2x−32x - 3 are polynomials. Lesson: algebra of continuous functions

3A limit only looks near the point

lim⁡x→cf(x)\lim_{x\to c} f(x) depends only on the values of ff at points close to c.c. If ff equals a polynomial pp on some open interval around c,c, then lim⁡x→cf(x)=p(c).\lim_{x\to c} f(x) = p(c). Basics: what a limit is

4One-sided limits with a small step hh

With h>0h > 0⁠: LHL at cc is lim⁡h→0f(c−h)\lim_{h\to0} f(c-h) and RHL is lim⁡h→0f(c+h).\lim_{h\to0} f(c+h). Basics: left-hand and right-hand limits

Solution

  1. Domain and pieces

    Each real xx satisfies exactly one of x≤2x \le 2 and x>2,x > 2, so ff is defined on all of R.\R. Split R\R into three parts: x<2,x < 2, x=2x = 2 and x>2.x > 2.

    Every point of the domain must be examined. Splitting this way separates the easy points (strictly inside a piece) from the one point where two formulas meet.
    Basics: piecewise functions

  2. Points c<2c < 2⁠: continuous

    Let c<2c < 2 and put δ=2−c>0.\delta = 2 - c > 0. Every xx with ∣x−c∣<δ|x - c| < \delta satisfies x<2,x < 2, so on that interval f(x)=2x+3.f(x) = 2x + 3. Hence

    lim⁡x→cf(x)=lim⁡x→c(2x+3)=2c+3=f(c),\lim_{x\to c} f(x) = \lim_{x\to c} (2x + 3) = 2c + 3 = f(c),

    and both one-sided limits equal this common limit. So ff is continuous at every c<2.c < 2.

    Near such a c,c, ff is the polynomial 2x+3,2x + 3, and a limit sees only nearby points (Basic 3). The polynomial is continuous by Example 14, so its limit at cc is its value 2c+3,2c + 3, which is also f(c).f(c).
    Every polynomial is continuous: Example 14.

  3. Points c>2c > 2⁠: continuous

    Let c>2c > 2 and put δ=c−2>0.\delta = c - 2 > 0. Every xx with ∣x−c∣<δ|x - c| < \delta satisfies x>2,x > 2, so there f(x)=2x−3,f(x) = 2x - 3, and

    lim⁡x→cf(x)=lim⁡x→c(2x−3)=2c−3=f(c).\lim_{x\to c} f(x) = \lim_{x\to c} (2x - 3) = 2c - 3 = f(c).

    So ff is continuous at every c>2.c > 2.

    The same argument as Step 2 with the other formula. The value f(c)=2c−3f(c) = 2c - 3 uses the second piece because c>2.c > 2.

  4. The junction c=2c = 2

    Value. 2≤2,2 \le 2, so the first piece applies: f(2)=2(2)+3=7.f(2) = 2(2) + 3 = 7.

    Left-hand limit. For h>0,h > 0, 2−h<2,2 - h < 2, so

    lim⁡x→2−f(x)=lim⁡h→0(2(2−h)+3)=lim⁡h→0(7−2h)=7.\begin{aligned} \lim_{x\to2^-} f(x) &= \lim_{h\to0} \bigl(2(2-h) + 3\bigr)\\ &= \lim_{h\to0} (7 - 2h) = 7. \end{aligned}

    Right-hand limit. For h>0,h > 0, 2+h>2,2 + h > 2, so

    lim⁡x→2+f(x)=lim⁡h→0(2(2+h)−3)=lim⁡h→0(1+2h)=1.\begin{aligned} \lim_{x\to2^+} f(x) &= \lim_{h\to0} \bigl(2(2+h) - 3\bigr)\\ &= \lim_{h\to0} (1 + 2h) = 1. \end{aligned}

    The one-sided limits are 7 and 1, which differ. So lim⁡x→2f(x)\lim_{x\to2} f(x) does not exist and ff is not continuous at x=2x = 2.

    x y −2 −1 1 2 3 4 −2 −1 1 2 3 4 5 6 7 8 9 O (2, 7) = f(2) (2, 1) not on the graph y = 2x + 3 y = 2x − 3
    Two parallel lines of slope 2. The left one, y=2x+3,y = 2x + 3, ends at the filled point (2,7)(2, 7)⁠; the right one, y=2x−3,y = 2x - 3, starts just after x=2x = 2 from the open circle (2,1).(2, 1). The graph drops by 6 at x=2.x = 2.
    The two formulas have the same slope but different constant terms, so they never meet: the gap between them is always (2x+3)−(2x−3)=6.(2x + 3) - (2x - 3) = 6. At the switch the graph has to jump by exactly that amount. No choice of f(2)f(2) could repair this, because the two one-sided limits themselves disagree.
    Compare Example 10, which has the same structure.

  5. Collect the answer

    ff is continuous at every c<2c < 2 (Step 2) and every c>2c > 2 (Step 3), and not at c=2c = 2 (Step 4). So x=2x = 2 is the only point of discontinuity.

    Steps 1 to 4 covered every real number exactly once, so no point has been missed.
Result

The only point of discontinuity is x=2x = 2: there the left-hand limit is 7 and the right-hand limit is 1. At every other real number ff is continuous. (Textbook answer: "Discontinuous at x=2x = 2".)

Check it with numbers

f(1.999)=2(1.999)+3=6.998f(1.999) = 2(1.999) + 3 = 6.998 and f(2)=7,f(2) = 7, but f(2.001)=2(2.001)−3=1.002.f(2.001) = 2(2.001) - 3 = 1.002. Moving across 2 by just 0.002 changes the value by about 6. Away from 2, for example at 0: f(−0.001)=2.998,f(-0.001) = 2.998, f(0)=3,f(0) = 3, f(0.001)=3.002,f(0.001) = 3.002, all close together.

Common mistakes
  1. Using 2x−32x - 3 for f(2).f(2). The condition x≤2x \le 2 includes 2, so f(2)=7.f(2) = 7.
  2. Only checking x=2.x = 2. A complete answer also explains why every other point is a point of continuity.
  3. Computing the limit at 2 from one formula only. At a junction each side uses its own piece; the two answers must be compared.
What to take away
  1. "Find all points of discontinuity": split at the junctions, argue continuity inside each piece, then test each junction with LHL, RHL and the value.
  2. Inside a piece, a polynomial formula is continuous (Example 14).
  3. Two lines with the same slope and different intercepts can never join up: a jump is unavoidable.

Every fact used, and where it comes from

FactUsed inSource
Continuity at cc⁠; point of discontinuitySteps 2–5Definition 1, Section 5.2 (Lesson: continuity at a point)
Every polynomial function is continuousSteps 2, 3Example 14
A limit depends only on values near the pointSteps 2, 3Class XI Limits (Basics: what a limit is)
One-sided limits as lim⁡h→0f(c∓h),\lim_{h\to0} f(c\mp h), h>0h > 0Step 4Class XI Limits (Basics: one-sided limits)
Same structure: a two-piece function with a jumpStep 4Example 10
Values in the check box and the graph coordinatesCheck; figureRecomputed in the site's verification script
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