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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q7: Discontinuities of a three-piece function with ∣x∣+3|x|+3

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)={∣x∣+3,if x≤−3−2x,if −3<x<36x+2,if x≥3f(x) = \begin{cases} |x| + 3, & \text{if } x \le -3 \\ -2x, & \text{if } -3 < x < 3 \\ 6x + 2, & \text{if } x \ge 3 \end{cases}

Exam answer

For x<−3x < -3⁠, f(x)=−x+3f(x) = -x + 3 (as ∣x∣=−x|x| = -x⁠); for −3<x<3-3 < x < 3⁠, f(x)=−2xf(x) = -2x⁠; for x>3x > 3⁠, f(x)=6x+2f(x) = 6x + 2⁠. These are polynomials, so ff is continuous at every xx other than −3-3 and 33⁠.

At x=−3x = -3⁠:

LHL=lim⁡h→0f(−3−h)=lim⁡h→0(∣−3−h∣+3)=lim⁡h→0(6+h)=6RHL=lim⁡h→0f(−3+h)=lim⁡h→0(−2(−3+h))=6\begin{aligned} \text{LHL} &= \lim_{h\to0} f(-3 - h) \\ &= \lim_{h\to0}\bigl(|-3 - h| + 3\bigr) \\ &= \lim_{h\to0}(6 + h) = 6 \\ \text{RHL} &= \lim_{h\to0} f(-3 + h) \\ &= \lim_{h\to0}\bigl(-2(-3 + h)\bigr) = 6 \end{aligned}

and f(−3)=∣−3∣+3=6f(-3) = |-3| + 3 = 6⁠. So LHL=RHL=f(−3)\text{LHL} = \text{RHL} = f(-3)⁠: ff is continuous at x=−3x = -3⁠.

At x=3x = 3⁠:

LHL=lim⁡h→0f(3−h)=lim⁡h→0(−2(3−h))=−6RHL=lim⁡h→0f(3+h)=lim⁡h→0(6(3+h)+2)=20\begin{aligned} \text{LHL} &= \lim_{h\to0} f(3 - h) \\ &= \lim_{h\to0}\bigl(-2(3 - h)\bigr) = -6 \\ \text{RHL} &= \lim_{h\to0} f(3 + h) \\ &= \lim_{h\to0}\bigl(6(3 + h) + 2\bigr) = 20 \end{aligned}

and f(3)=6(3)+2=20f(3) = 6(3) + 2 = 20⁠. Since LHL≠RHL\text{LHL} \ne \text{RHL}⁠, ff is not continuous at x=3x = 3⁠.

Hence x=3x = 3 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

There are two junctions, x=−3x = -3 and x=3,x = 3, and three open pieces between them. Inside each piece ff is a polynomial (once ∣x∣|x| is rewritten as −x-x for negative xx⁠), so it is continuous there. Each junction is then tested on its own. At −3-3 the neighbouring pieces happen to meet at the same height, 6, so there is no break. At 33 they do not (−6-6 on the left, 20 on the right), so there is a jump.

Given, to find, and the plan
Given
A three-piece function defined for every real x.x.
To find
Every point of the domain at which ff is not continuous.
Plan
(1) Domain; simplify ∣x∣|x| on its piece. (2) Points inside the three pieces. (3) Junction x=−3.x = -3. (4) Junction x=3.x = 3. (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; otherwise cc is a point of discontinuity (Definition 1, Section 5.2). Lesson: continuity at a point

2Absolute value of a negative number

∣x∣=−x|x| = -x when x<0.x < 0. For example ∣−3∣=3\lvert -3\rvert = 3 and ∣−3.5∣=3.5.\lvert -3.5\rvert = 3.5. So for x≤−3,x \le -3, ∣x∣+3=−x+3.|x| + 3 = -x + 3. Basics: piecewise functions

3Polynomials are continuous; limits are local

Every polynomial is continuous (Example 14). If ff equals a polynomial pp on an 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 remove the modulus

    Every real xx satisfies exactly one of x≤−3,x \le -3, −3<x<3,-3 < x < 3, x≥3,x \ge 3, so ff is defined on all of R.\R. On the first piece x≤−3<0,x \le -3 < 0, so ∣x∣=−x|x| = -x and

    f(x)=−x+3for x≤−3.f(x) = -x + 3 \quad \text{for } x \le -3.

    The real line splits into: x<−3,x < -3, the point −3,-3, −3<x<3,-3 < x < 3, the point 3,3, and x>3.x > 3.

    Rewriting ∣x∣+3|x| + 3 as −x+3-x + 3 on its piece turns all three formulas into polynomials, so the same argument works inside every piece. The conditions overlap nowhere and leave no gaps, so each xx has exactly one value.
    ∣x∣=−x|x| = -x for x<0.x < 0. Basics: piecewise functions

  2. Points inside the three pieces: continuous

    • If c<−3c < -3⁠: for ∣x−c∣<−3−c|x - c| < -3 - c we have x<−3,x < -3, so f(x)=−x+3f(x) = -x + 3 near c,c, and lim⁡x→cf(x)=−c+3=f(c).\lim_{x\to c} f(x) = -c + 3 = f(c).
    • If −3<c<3-3 < c < 3⁠: for xx closer to cc than both −3-3 and 33 are, f(x)=−2x,f(x) = -2x, so lim⁡x→cf(x)=−2c=f(c).\lim_{x\to c} f(x) = -2c = f(c).
    • If c>3c > 3⁠: for ∣x−c∣<c−3|x - c| < c - 3 we have x>3,x > 3, so f(x)=6x+2f(x) = 6x + 2 near c,c, and lim⁡x→cf(x)=6c+2=f(c).\lim_{x\to c} f(x) = 6c + 2 = f(c).

    So ff is continuous at every point other than −3-3 and 3.3.

    At a point strictly inside a piece, there is a small interval around it that stays inside the same piece. On that interval ff is a polynomial, which is continuous (Example 14), and a limit only looks at that interval.
    Every polynomial is continuous: Example 14.

  3. The junction x=−3x = -3⁠: continuous

    Value. −3≤−3,-3 \le -3, so the first piece applies: f(−3)=∣−3∣+3=6.f(-3) = \lvert -3\rvert + 3 = 6.

    Left-hand limit. For h>0,h > 0, −3−h<−3,-3 - h < -3, so the first piece applies:

    lim⁡x→−3−f(x)=lim⁡h→0(∣−3−h∣+3)=lim⁡h→0(3+h+3)=lim⁡h→0(6+h)=6.\begin{aligned} \lim_{x\to-3^-} f(x) &= \lim_{h\to0} \bigl(\lvert -3-h\rvert + 3\bigr)\\ &= \lim_{h\to0} (3 + h + 3)\\ &= \lim_{h\to0} (6 + h) = 6. \end{aligned}

    Right-hand limit. For 0<h<6,0 < h < 6, −3<−3+h<3,-3 < -3 + h < 3, so the middle piece applies:

    lim⁡x→−3+f(x)=lim⁡h→0(−2(−3+h))=lim⁡h→0(6−2h)=6.\begin{aligned} \lim_{x\to-3^+} f(x) &= \lim_{h\to0} \bigl(-2(-3+h)\bigr)\\ &= \lim_{h\to0} (6 - 2h) = 6. \end{aligned}

    All three numbers are 6, so ff is continuous at x=−3x = -3.

    ∣−3−h∣=3+h\lvert -3 - h\rvert = 3 + h because −3−h-3 - h is negative and its distance from 0 is 3+h.3 + h. The two neighbouring formulas give the same height at −3-3⁠: ∣−3∣+3=6\lvert -3\rvert + 3 = 6 and −2(−3)=6.-2(-3) = 6. That is why the pieces join up there, as the graph shows. A junction is only a candidate; it still has to be tested.
    Basics: left-hand and right-hand limits

  4. The junction x=3x = 3⁠: a jump

    Value. 3≥3,3 \ge 3, so the third piece applies: f(3)=6(3)+2=20.f(3) = 6(3) + 2 = 20.

    Left-hand limit. For 0<h<6,0 < h < 6, −3<3−h<3,-3 < 3 - h < 3, so the middle piece applies:

    lim⁡x→3−f(x)=lim⁡h→0(−2(3−h))=lim⁡h→0(−6+2h)=−6.\begin{aligned} \lim_{x\to3^-} f(x) &= \lim_{h\to0} \bigl(-2(3-h)\bigr)\\ &= \lim_{h\to0} (-6 + 2h) = -6. \end{aligned}

    Right-hand limit. For h>0,h > 0, 3+h>3,3 + h > 3, so the third piece applies:

    lim⁡x→3+f(x)=lim⁡h→0(6(3+h)+2)=lim⁡h→0(20+6h)=20.\begin{aligned} \lim_{x\to3^+} f(x) &= \lim_{h\to0} \bigl(6(3+h) + 2\bigr)\\ &= \lim_{h\to0} (20 + 6h) = 20. \end{aligned}

    The one-sided limits −6-6 and 2020 differ, so ff is not continuous at x=3x = 3.

    x y −7 −6 −5 −4 −3 −2 −1 1 2 3 4 −8 −4 4 8 12 16 20 24 28 32 O (−3, 6): pieces meet (3, −6): not on the graph (3, 20) = f(3) y = |x| + 3 y = −2x y = 6x + 2
    The three pieces of f.f. At x=−3x = -3 the pieces y=∣x∣+3y = |x| + 3 and y=−2xy = -2x meet at the filled point (−3,6)(-3, 6)⁠: no break. At x=3x = 3 the middle piece ends at the open circle (3,−6)(3, -6) and the last piece starts at the filled point (3,20)(3, 20)⁠: a jump of 26. (The two axes use different scales; read values from the tick labels.)
    The value f(3)=20f(3) = 20 agrees with the right-hand limit, but the left-hand limit is −6.-6. Continuity needs both sides to agree with f(3),f(3), so it fails.

  5. Collect the answer

    ff is continuous inside each piece (Step 2) and at x=−3x = -3 (Step 3), but not at x=3x = 3 (Step 4). So x=3x = 3 is the only point of discontinuity.

    Steps 2 to 4 cover every real number, so nothing is missed. Notice that one junction turned out continuous and the other did not: junctions must be checked individually.
Result

The only point of discontinuity is x=3x = 3 (left-hand limit −6,-6, right-hand limit 20). At x=−3x = -3 the left-hand limit, the right-hand limit and f(−3)f(-3) all equal 6, so ff is continuous there, and it is continuous at every other real number too. (Textbook answer: "Discontinuous at x=3x = 3".)

Check it with numbers
  • Near −3-3⁠: f(−3.001)=3.001+3=6.001,f(-3.001) = 3.001 + 3 = 6.001, f(−3)=6,f(-3) = 6, f(−2.999)=−2(−2.999)=5.998.f(-2.999) = -2(-2.999) = 5.998. All close to 6.
  • Near 3: f(2.999)=−5.998f(2.999) = -5.998 but f(3)=20f(3) = 20 and f(3.001)=20.006.f(3.001) = 20.006. A gap of about 26 that does not shrink.
Common mistakes
  1. ∣−3−h∣=−3−h.\lvert -3 - h\rvert = -3 - h. An absolute value is never negative; since −3−h<0,-3 - h < 0, ∣−3−h∣=3+h.\lvert -3 - h\rvert = 3 + h.
  2. Assuming every junction is a discontinuity. At −3-3 the pieces meet, so ff is continuous there. Test each junction.
  3. Taking the value at x=3x = 3 from the middle piece. The middle condition is strict (x<3x < 3⁠); x=3x = 3 belongs to x≥3,x \ge 3, so f(3)=20.f(3) = 20.
  4. Using the right piece for the right-hand limit at −3-3 without checking. Points just right of −3-3 satisfy −3<x<3,-3 < x < 3, so the middle formula −2x-2x applies.
What to take away
  1. With several pieces, list every junction and test each one separately; inside the pieces, polynomial formulas are continuous.
  2. Remove a modulus on its piece using the sign of xx there (here ∣x∣=−x|x| = -x for x≤−3x \le -3⁠).
  3. A junction is continuous exactly when the neighbouring formulas give the same height there and ff takes that height.

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)
∣x∣=−x|x| = -x for x<0x < 0Steps 1, 3Definition of modulus (Basics: piecewise functions)
Every polynomial function is continuousStep 2Example 14
A limit depends only on values near the pointStep 2Class 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 > 0Steps 3, 4Class XI Limits (Basics: one-sided limits)
Values in the check box and the graph coordinatesCheck; figureRecomputed in the site's verification script
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