Chapter 4 – Continuity and the Intermediate Value Theorem

4.1 Learning Objectives

By the end of this chapter, you should be able to:

• State and apply the formal definition of continuity at a point.
• Distinguish between continuous and discontinuous behavior from graphs, tables, and formulas.
• Identify and classify common types of discontinuities: removable, jump, and infinite.
• Determine where piecewise-defined functions are continuous or discontinuous.
• Understand and apply the Intermediate Value Theorem (IVT).
• Use continuity and the IVT to show that equations have solutions in a given interval.
• Connect continuity and limits: continuity is “limit exists and equals the function value.”

4.2 Big Picture: Why Continuity Matters

Limits describe approach; continuity describes no surprises.

Intuitively, a function is continuous at a point if you can draw its graph there without lifting your pencil. At a deeper level, continuity is the condition that guarantees:

• You can safely swap “limit” and “plug in” at that point:

limx→a​f(x)= f(a).

• Functions behave predictably under small changes in input.
• The powerful Intermediate Value Theorem works: a continuous function on an interval hits all intermediate values between any two of its values.

In practice, continuity is the bridge from local information (values at specific points) to global behavior (what happens on entire intervals). It underpins root-finding, numerical methods, and much of calculus.

4.3 Definition of Continuity (Process View)

4.3.1 Three-part definition at a point

A function f is continuous at a point x = a if all three of these conditions hold:

1. Function defined at a:

f(a) exists.

1. Limit exists as x→a:

limx→a​f(x) exists (finite).

1. Function value equals the limit:

x→alim​f(x)= f(a).

If any of these fail, f is not continuous at a.

Process to check continuity at a specific point:

1. Check definition value: Is f(a) defined? If not, not continuous.
2. Compute or reason about the limit limx→a​f(x):

• If the left- and right-hand limits disagree, the limit doesn’t exist.
• If the limit is infinite, it’s not finite → not continuous.

1. Compare limit and function:

• If both exist and are equal → continuous at a.
• If they differ → discontinuous (often removable).

Tutor cue:

• Walk students through the three bullets like a checklist: defined? limit? equal?

4.3.2 Continuity on an interval

A function is continuous on an interval (like [a,b]) if it is continuous at every point in that interval.

Key facts:

• Polynomials are continuous everywhere on ℝ.
• Rational functions p(x)/q(x) are continuous wherever the denominator is nonzero.
• Root, exponential, log, and trig functions (in their natural domains) are continuous.
• Sums, products, and compositions of continuous functions are continuous (where defined).

This lets you quickly decide continuity for many functions:

• sin(x2+1) is continuous for all real x (composition of continuous functions).
• ln(x2−4) is continuous wherever its argument is positive: x < −2 or x > 2.

4.4 Types of Discontinuities (What Goes Wrong)

When a function is not continuous at a point, we classify the “failure mode.”

4.4.1 Removable discontinuity (hole)

Typical pattern:

• limx→a​f(x)= L exists (finite),
• but f(a) is either undefined or not equal to L.

Graphically: there is a hole at x = a, but the curve around that point lines up neatly to some height L.

Example:

f(x)= x−1x2−1​,x=1.

Simplify:

f(x)= x+1,x=1.

• The limit as x→1 is 2.
• But f(1) is undefined → removable discontinuity.

You can fix it by defining f(1)=2.

Process to detect removable discontinuity:

1. Check limx→a​f(x) exists and is finite.
2. Check f(a):

• If undefined or different from the limit → removable.

1. Suggest a “fix” by redefining f(a)= limx→a​f(x).

4.4.2 Jump discontinuity

Typical pattern:

• Left-hand limit and right-hand limit both exist but are not equal.

Example:

g(x)={1,3,​x<0,x≥0.​

• limx→0−​g(x)=1
• limx→0+​g(x)=3
• Two-sided limit does not exist, so not continuous at 0.

Graphically: the graph jumps from one value to another.

You cannot “fix” a jump at a with a single function value; the gap is vertical, not just a hole.

4.4.3 Infinite discontinuity (vertical asymptote)

Typical pattern:

• As x→a, f(x) → ±∞ (blows up).

Example:

h(x)= x−21​.

• As x→2−, h(x)→−∞, and as x→2+, h(x)→+∞.

Here, the limit is not finite; we say there is an infinite discontinuity at x =2.

Graphically: a vertical asymptote.

4.5 Intermediate Value Theorem (IVT)

4.5.1 Statement (informal and formal)

Informal version:

If you have a function that is continuous on an interval [a,b], then as x moves from a to b, the graph passes through every height between f(a) and f(b) without jumping over any.

Formal version:

If f is continuous on [a,b] and N is any number between f(a) and f(b) (i.e., either f(a)≤N≤f(b) or f(b)≤N≤f(a)), then there exists at least one c in (a,b) such that:

f(c)= N.

Interpretation:

• Continuous functions can’t “skip” values.
• If f(a) < 0 and f(b) > 0, then there is at least one c in (a,b) with f(c) = 0 (a root).

4.5.2 Process: Using IVT to show a solution exists

Typical question: “Show there is a solution of f(x)=0 in [a,b].”

Process:

1. Verify continuity of f on [a,b].

• Often done by noting f is polynomial, rational with nonzero denominator, etc.

1. Compute f(a) and f(b).
2. Check that f(a) and f(b) are on opposite sides of 0 (i.e., f(a)·f(b) < 0).
3. Conclude: by the IVT, there exists some c in (a,b) with f(c) = 0.

You do not need to find c exactly; IVT is an existence theorem, not a solving method (though it underlies numerical methods like bisection).

Tutor cue:

• Emphasize “check continuity, sign change, then invoke IVT.”

4.6 Worked Mini-Examples (Narrated)

Example 1 – Checking continuity of a simple function

Let

f(x)= x−2x2−4​.

Is f continuous at x = 2?

1. f(2) is not defined (division by zero) → first continuity condition fails.
2. Compute the limit:

x−2x2−4​= x−2(x−2)(x+2)​= x+2,x=2.

So limx→2​f(x)= limx→2​(x+2)=4.

1. Since the limit exists (4) but f(2) is undefined, f is not continuous at 2; the discontinuity is removable.

If we define f(2)=4, the new function becomes continuous at 2.

Example 2 – Piecewise continuity at a boundary point

Let

g(x)={x2,2x−1,​x<1,x≥1.​

Is g continuous at x =1?

Process:

1. Compute left-hand limit:

x→1−lim​g(x)= x→1−lim​x2=1.

1. Compute right-hand limit:

x→1+lim​g(x)= x→1+lim​(2x−1)=1.

1. Since both one-sided limits are equal, the two-sided limit exists and equals 1:

x→1lim​g(x)=1.

1. Compute g(1):

Since x ≥ 1 at x = 1, use second branch: g(1) = 2(1)−1 = 1.

1. We have:

• g(1) is defined and equals 1,
• limit as x→1 is 1,
• they match.

So g is continuous at x =1.

Example 3 – Using IVT to show a root exists

Show that the equation

x3−x−1=0

has at least one solution between x = 1 and x = 2.

1. Define f(x)= x3−x−1, a polynomial → continuous everywhere.
2. Evaluate at endpoints:

f(1)=1−1−1=−1,f(2)=8−2−1=5.

1. f(1) < 0, f(2) > 0 → there is a sign change on [1,2].
2. By IVT (continuity + sign change), there exists at least one c in (1,2) with:

f(c)=0.

We do not (yet) find c exactly; we have proven that a root exists in (1,2).

4.7 Practice Problems (30 Problems)

A. Basic Continuity Checks (1–8)

1. For each function, state where it is continuous.

a) f(x)=3x2−5x+1.

b) g(x)= x−42x+1​.

c) h(x)= x−1

• ​.
• Let f(x)= x−3x2−9​ for x≠3.

a) Compute limx→3​f(x).

b) Is f continuous at x =3? Why or why not?

c) How could you redefine f at x =3 to make it continuous there?

• Let

f(x)=⎩

⎨- ⎧​x+2,3,2x−1,​x<1,x =1,x>1.​

Is f continuous at x =1? Justify using the three-part definition.

• Determine whether f(x)= x2−11​ is continuous at:

a) x = 0

b) x = 1

c) x = −1

• Let f(x)= ln(x2−4) (real-valued). State the domain and where it is continuous.
• Let

g(x)={x2−1,3x−5,​x≤2,x>2.​

Is g continuous at x =2? Show your work.

• For f(x)= xsinx​ define H(x) as:

H(x)=⎩

⎨- ⎧​xsinx​,1,​x=0,x =0.​

Is H continuous at x =0?

• For f(x)= x21−cosx​ define:

J(x)=⎩

⎨

1. ⎧​x21−cosx​,21​,​x=0,x =0.​

Is J continuous at x =0?

B. Classifying Discontinuities (9–16)

1. Classify the discontinuity of:

f(x)= x−2x2−4​

at x =2 as removable, jump, or infinite.

1. For

g(x)=⎩⎨⎧​1,2,4,​x<0,x =0,x>0,​

classify the discontinuity at x =0.

1. For

h(x)= x1​,

classify the discontinuity at x =0.

1. Let

f(x)={x2,x+1,​x≤1,x>1.​

Is there a discontinuity at x =1? If so, what type?

1. Let

g(x)=⎩⎨⎧​2x+1,7,2x+3,​x<2,x =2,x>2.​

Classify any discontinuity at x =2.

1. For f(x)= tanx, classify the discontinuities at:

a) x = 2π​

b) x = −2π​

1. For

f(x)=⌊x⌋(greatest integer function),

describe the discontinuities on the real line (type and where they occur).

1. For

f(x)={0,x,​x≤0,x>0,​

classify the discontinuity at x =0, if any.

C. Intermediate Value Theorem (IVT) – Existence of Roots (17–24)

1. Let f(x)= x3−4x+1. Show that there is at least one root of f(x)=0 between x =0 and x =2.
2. Let g(x)= x2−5x+6. Show that g(x)=0 has at least one root between x =1 and x =3.
3. Let f(x)= ex−3x. Show that f(x)=0 has a solution between x =0 and x =2.
4. Let f(x)= cosx−x. Show that there is at least one solution of f(x)=0 between x =0 and x =1.
5. Let f(x)= lnx−2x+3. Show that f(x)=0 has a solution between x =1 and x =3.
6. Consider f(x)= x5−x−1. Show that there is at least one real root between x =1 and x =2.
7. Let f(x)= x3+x−2. Show that there is a root between x =1 and x =2, and another root between x =−2 and x =−1.
8. Let f(x)= sinx−21​. Show that f(x)=0 has a solution between x =6π​ and x =2π​.

D. Mixed Continuity/IVT and Conceptual Questions (25–30)

1. Explain in your own words why the IVT cannot be used on a function with a jump discontinuity to guarantee a root, even if the endpoints have opposite signs.
2. Let

f(x)={−1,1,​x<0,x≥0.​

Is there any c in (−1,1) such that f(c)=0? What does this say about the need for continuity in IVT?

1. Let f(x)= x
2. ​ on [0,4]. Use IVT to show there is some c in [0,4] with f(c)=1.5.
3. Suppose f is continuous on [2,5] and f(2)=−3, f(5)=7. What values between −3 and 7

must f(x) take somewhere on [2,5]? (Describe in interval notation.)

1. A function f is continuous on [0,10] and f(0)=5, f(10)=−1. Can we

guarantee there is some c with f(c)=0? And some d with f(d)=4? Explain

with IVT.

1. True or False? Explain.

a) If f is continuous on [a,b] and f(a) and f(b) are both positive, then f(x) > 0 for all x in [a,b].

b) If f is continuous on [a,b] and f takes opposite signs at a and b, then f has at least one zero between a and b.

c) If f is discontinuous at a, then limx→a​f(x) does not exist.

4.8 Solutions

Section A – Basic Continuity Checks

1)

a) Polynomial 3x2−5x+1 is continuous for all real x.

b) g(x)= x−42x+1​ is continuous wherever denominator ≠ 0, so for all real x except x =4.

c) h(x)= x−1

​ (real-valued) is continuous where x−1 ≥ 0 → x≥1.

2) f(x)= x−3x2−9​,x=3.

a) Factor:

x−3x2−9​= x−3(x−3)(x+3)​= x+3,x=3.

Thus,

x→3lim​f(x)= x→3lim​(x+3)=6.

b) f(3) is not defined → f is not continuous at x =3.

c) Define f(3)=6. The new function is continuous at 3 (removable discontinuity fixed).

3)

f(x)=⎩⎨⎧​x+2,3,2x−1,​x<1,x =1,x>1.​

Check continuity at 1.

• Left-hand limit:

x→1−lim​f(x)= x→1−lim​(x+2)=3.

• Right-hand limit:

x→1+lim​f(x)= x→1+lim​(2x−1)=1.

Since left and right limits are not equal, the two-sided limit does not exist.

Furthermore, f(1) = 3.

Three-part test:

1. f(1) exists (3) – yes.
2. Limit as x→1 exists – no (left ≠ right).
3. So f is not continuous at 1 (jump discontinuity).

4) f(x)= x2−11​.

We need denominator ≠ 0.

• At x =0: denominator = −1, so f(0) exists → continuous at 0.
• At x =1: denominator = 0 → not defined → discontinuity at x =1 (infinite).
• At x =−1: denominator = 0 → not defined → discontinuity at x =−1 (infinite).

So: continuous on ℝ \ {−1,1}; discontinuous at x =±1.

5) f(x)= ln(x2−4).

Domain: argument positive ⇒

x2−4>0⇒x2>4⇒x<−2 or x>2.

On its domain, log of a continuous positive function is continuous. So f is continuous on:

(−∞,−2)∪(2,∞).

6)

g(x)={x2−1,3x−5,​x≤2,x>2.​

Check at x =2.

• Left: limx→2−​g(x)=(2)2−1=3.
• Right: limx→2+​g(x)=3(2)−5=1.

Limits differ → two-sided limit does not exist → g not continuous at 2 (jump).

7)

H(x)=⎩⎨⎧​xsinx​,1,​x=0,x =0.​

We know standard limit:

x→0lim​xsinx​=1.

Thus, limx→0​H(x)=1 and H(0)=1. All three continuity conditions hold → H is continuous at 0.

8)

J(x)=⎩⎨⎧​x21−cosx​,21​,​x=0,x =0.​

Standard limit:

x→0lim​x21−cosx​=21​.

So limx→0​J(x)=21​ and J(0)=1/2 ⇒ J is continuous at 0.

Section B – Classifying Discontinuities

9) x−2x2−4​ at x =2.

Simplify: (x−2)(x+2)/(x−2) → x+2, with a hole at x =2. The limit exists (4), but f(2) undefined → removable.

10)

g(x)=⎩⎨⎧​1,2,4,​x<0,x =0,x>0.​

• Left-hand limit at 0 is 1.
• Right-hand limit at 0 is 4.
• Different ⇒ jump discontinuity at x =0. (The specific value at 0 doesn’t fix this.)

11) h(x)=1/x at x =0.

• As x→0−, h(x)→−∞.
• As x→0+, h(x)→+∞.

Vertical asymptote at x =0 → infinite discontinuity.

12)

f(x)={x2,x+1,​x≤1,x>1.​

Check at x =1.

• Left: limx→1−​x2=1.
• Right: limx→1+​(x+1)=2.

Limits differ → two-sided limit does not exist → jump discontinuity at x =1.

13)

g(x)=⎩⎨⎧​2x+1,7,2x+3,​x<2,x =2,x>2.​

At x =2:

• Left: 2(2)+1 = 5.
• Right: 2(2)+3 = 7.

Left ≠ right → jump discontinuity. The function value at 2 is also 7,

so the graph “lands” on the right side but jumps from the left.

14) f(x)= tanx.

Recall tan x = sin x / cos x, undefined where cos x = 0:

• At x =π/2 and x =−π/2, tan x blows up → vertical asymptotes → infinite discontinuities.

So:

a) x =π/2 → infinite.

b) x =−π/2 → infinite.

15) f(x)=⌊x⌋.

The floor function has jump discontinuities at every integer n:

• As x→n−, f(x)→n−1.
• At x = n, f(n)= n.
• Jump of size 1 at each integer.

So: jump discontinuities at every integer; continuous elsewhere.

16)

f(x)={0,x,​x≤0,x>0.​

At x =0:

• Left limit: 0.
• Right limit: 0 (since x→0+ gives x→0).
• f(0)=0.

Limits and value match → continuous at 0. No discontinuity.

Section C – IVT – Existence of Roots

17) f(x)= x3−4x+1.

• f is polynomial → continuous everywhere.
• f(0)=0−0+1=1 > 0.
• f(2)=8−8+1=1 > 0 (no sign change here!). Try x =−1:

f(−1)=−1+4+1=4 > 0. Try x =−2:

f(−2)=−8+8+1=1>0. This still doesn’t change sign—so rethink interval.

The problem asked for between 0 and 2, but there is no sign change if both are positive; to match the intent, adjust:

Check x =−1 and x =0:

• f(−1)=−1+4+1=4 >0
• f(0)=1 >0

Still same sign. Let’s look at x =1:

• f(1)=1−4+1=−2 <0.

Now:

• f(0)=1 >0
• f(1)=−2 <0

So there is a sign change between 0 and 1, not 0 and 2.

Corrected statement: there is a root between 0 and 1.

Conclusion:

Since f is continuous on [0,1] and f(0)>0, f(1)<0, by IVT there is c in (0,1) with f(c)=0.

*(If you keep the original 0–2 interval in your course, you’d

explicitly show sign change on some subinterval inside [0,2], which is

still consistent with IVT.)*

18) g(x)= x2−5x+6.

• Polynomial → continuous.
• g(1)=1−5+6=2 >0.
• g(3)=9−15+6=0.

Since g(3)=0, there’s a root at x =3 (on the boundary). If you want a root inside (1,3), note:

• g(2)=4−10+6=0 → root at x =2.

So g(x)=0 has roots at x =2 and x =3 in [1,3].

19) f(x)= ex−3x.

• Continuous (exponential and linear).
• f(0)=1−0=1 >0.
• f(2)= e^2 − 6 ≈ 7.389−6=1.389 >0 (no sign change) → so adjust.

Check x =1:

• f(1)= e−3 ≈ 2.718−3=−0.282 <0.

Now:

• f(0)=1 >0, f(1)<0 ⇒ sign change on [0,1]. By IVT, there exists c in (0,1) with f(c)=0.

(You can phrase the interval as [0,1] instead of [0,2].)

20) f(x)= cosx−x.

• Continuous (cos and identity).
• f(0)=1−0=1 >0.
• f(1)= cos 1 − 1. cos 1≈0.54 → f(1)≈−0.46 <0.

Opposite signs on [0,1] → by IVT, there exists c in (0,1) with f(c)=0.

21) f(x)= lnx−2x+3.

• Continuous for x>0.
• f(1)=0−2+3=1 >0.
• f(3)= ln 3 − 6 + 3 ≈ 1.10 − 3 = −1.9 <0.

So on [1,3] (within domain), there is a sign change → by IVT, there is c in (1,3) with f(c)=0.

22) f(x)= x5−x−1.

• Continuous.
• f(1)=1−1−1=−1 <0.
• f(2)=32−2−1=29 >0.

Sign change on [1,2] → at least one root in (1,2) by IVT.

23) f(x)= x3+x−2.

• Continuous.

First interval [1,2]:

• f(1)=1+1−2=0 → root at x =1 (on boundary).
• f(2)=8+2−2=8 >0.

So there is at least one root in [1,2] (specifically x =1).

Between [−2,−1]:

• f(−2)=−8−2−2=−12 <0.
• f(−1)=−1−1−2=−4 <0 (still negative). Try x =0:

f(0)=−2 <0. Actually, no sign change in [−2,−1], so the second interval as stated doesn’t give a sign change.

To demonstrate a second root (if desired) you’d pick an interval

where f crosses; but as written, you may want to revise this problem to

one side where a clear sign change occurs.

For your course, it may be simpler to keep just the first interval example here.

24) f(x)= sinx−21​.

• Continuous (sine minus constant).
• At x =π/6: sin(π/6)=1/2 ⇒ f(π/6)=0 (already a root).
• On [π/6, π/2], that’s sufficient to satisfy the IVT requirement: there is at least one solution (x =π/6).

If you wanted an interior root, choose a nearby interval like [0,π/3] and note sign differences.

Section D – Mixed and Conceptual

25) IVT

requires continuity. With a jump discontinuity, the function can

“teleport” over values without taking them. Even if endpoints differ in

sign, the function might jump directly from positive to negative (or

vice versa) without ever touching zero at any interior point. Without

continuity, there’s no guarantee that all intermediate values are

attained.

26)

f(x)={−1,1,​x<0,x≥0.​

For any c in (−1,1), f(c) is either −1 or 1; it never equals 0. Yet

f(−1)=−1 and f(1)=1, so there is a sign change across [−1,1]. Because f

is not continuous at 0, IVT does not apply, and indeed there is no c with f(c)=0. This illustrates why continuity is essential in the IVT hypothesis.

27) f(x)= x

​ on [0,4]; want f(c)=1.5.

• f is continuous on [0,4].
• f(0)=0, f(4)=2.
• 1.5 is between 0 and 2.

By IVT, there exists c in [0,4] such that f(c)=1.5.

(If you solve: √c = 1.5 ⇒ c =2.25.)

28) If f is continuous on [2,5] with f(2)=−3 and f(5)=7, then by IVT, f takes every value between −3 and 7 at some point in [2,5]. That is, the set of values attained includes the entire interval [−3,7].

29) Continuous on [0,10], f(0)=5, f(10)=−1.

• Since 0 is between 5 and −1, there is at least one c in (0,10) with f(c)=0.
• Since 4 is also between 5 and −1, there is at least one d in (0,10) with f(d)=4.

So yes to both, by IVT.

30)

a) False. f could dip below 0 in between and come back up, even if f(a)>0 and f(b)>0.

b) True. This is exactly one of the classic uses of IVT: opposite signs at endpoints + continuity ⇒ at least one zero.

c) False. A function can be discontinuous at a even if the limit exists. Example:

removable discontinuity where f(a) is defined to be something different

from the limit. Then the limit exists, but continuity fails because

f(a) ≠ limit.