Intermediate Value Theorem Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Intermediate Value Theorem.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

If ff is continuous on the closed interval [a,b][a, b] and NN is any value between f(a)f(a) and f(b)f(b), then there exists at least one cc in (a,b)(a, b) such that f(c)=Nf(c) = N.

A continuous function can't skip values. If you start below a line and end above it, you must cross it somewhere. It's like driving from sea level to a mountaintopβ€”you pass through every elevation in between.

Read the full concept explanation β†’

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: If a continuous function starts below a level and ends above it, it must hit that level somewhere between.

Common stuck point: The procedure for intermediate value theorem is the easy part; the trap is applying IVT without checking continuity. Asking "Is the function continuous on a closed interval, and am I asked to show a value between the endpoints is attained somewhere inside?" first is what keeps a correct-looking calculation from being attached to the wrong concept.

Sense of Study hint: Ask: Is the function continuous on a closed interval, and am I asked to show a value between the endpoints is attained somewhere inside?

Worked Examples

Example 1

easy
Show that f(x)=x3βˆ’xβˆ’1f(x) = x^3 - x - 1 has a root in the interval (1,2)(1, 2).

Answer

By IVT, ff has at least one root in (1,2)(1, 2).

First step

1
ff is a polynomial, hence continuous on [1,2][1, 2].

Full solution

  1. 2
    f(1)=1βˆ’1βˆ’1=βˆ’1<0f(1) = 1 - 1 - 1 = -1 < 0.
  2. 3
    f(2)=8βˆ’2βˆ’1=5>0f(2) = 8 - 2 - 1 = 5 > 0.
  3. 4
    Since ff is continuous, f(1)<0<f(2)f(1) < 0 < f(2), by the IVT there exists c∈(1,2)c \in (1,2) with f(c)=0f(c) = 0.
The IVT requires continuity and a sign change. Polynomials are continuous everywhere, so verifying the sign change at the endpoints is sufficient.

Example 2

medium
Use the IVT to show that cos⁑x=x\cos x = x has a solution in (0,1)(0, 1).

Example 3

easy
Show g(x)=x3+2xβˆ’5g(x)=x^3+2x-5 has a root in (1,2)(1,2).

Example 4

medium
Show ln⁑x=1βˆ’x\ln x = 1-x has a solution in (1,e)(1,e).

Example 5

medium
Show the equation x=cos⁑(x)x = \cos(x) has a solution in [0,1][0,1].

Example 6

hard
Show that every continuous function f:[0,1]β†’[0,1]f:[0,1] \to [0,1] has a fixed point.

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Show that h(x)=x2βˆ’2h(x) = x^2 - 2 has a root in (1,2)(1, 2).

Example 2

medium
Can IVT be applied to f(x)=1xf(x) = \frac{1}{x} on [βˆ’1,1][-1, 1] to conclude it attains the value 0? Explain.

Example 3

easy
Does f(x)=x2f(x) = x^2 take the value 2 somewhere on [0,2][0, 2]? (It is continuous.)

Example 4

easy
Given ff continuous with f(1)=βˆ’3f(1) = -3 and f(4)=5f(4) = 5, must ff have a root in (1,4)(1,4)?

Example 5

easy
State the two hypotheses required to apply the IVT on [a,b][a,b].

Example 6

easy
Can IVT be applied to f(x)=1xf(x) = \frac{1}{x} on [βˆ’1,1][-1, 1]?

Example 7

easy
A continuous function has f(0)=1f(0) = 1 and f(3)=7f(3) = 7. Is there a cc with f(c)=4f(c) = 4?

Example 8

easy
Does f(x)=cos⁑xf(x) = \cos x equal 0 somewhere on [0,2][0, 2]?

Example 9

easy
Why must ff be continuous on the CLOSED interval [a,b][a,b] for IVT?

Example 10

easy
If ff is continuous and f(2)=f(5)=3f(2) = f(5) = 3, does IVT guarantee ff takes the value 10 on [2,5][2,5]?

Example 11

medium
Show f(x)=x3+xβˆ’1f(x) = x^3 + x - 1 has a root in (0,1)(0, 1).

Example 12

medium
Show cos⁑x=x\cos x = x has a solution in (0,1)(0, 1).

Example 13

medium
Show f(x)=x5βˆ’2x+3f(x) = x^5 - 2x + 3 has a real root.

Example 14

medium
A continuous temperature is βˆ’2∘-2^\circ at 6am and 10∘10^\circ at noon. Was it exactly 0∘0^\circ at some time?

Example 15

medium
Show f(x)=ex+x=2f(x) = e^x + x = 2 has exactly one solution, locating an interval.

Example 16

medium
A continuous ff maps [0,1][0,1] into [0,1][0,1]. Show ff has a fixed point (f(c)=cf(c)=c).

Example 17

medium
Show tan⁑x=x\tan x = x has a solution in (Ο€,3Ο€2)\left(\pi, \frac{3\pi}{2}\right)... wait, why must we avoid Ο€2\frac{\pi}{2} intervals?

Example 18

medium
Given continuous ff with f(1)=2,f(2)=βˆ’1,f(3)=4f(1)=2, f(2)=-1, f(3)=4, what is the minimum number of roots IVT guarantees on [1,3][1,3]?

Example 19

medium
Prove every odd-degree polynomial p(x)p(x) has at least one real root.

Example 20

challenge
A hiker climbs a trail 8am-2pm, descends the same trail next day 8am-2pm. Prove there is a clock time at the same spot both days.

Example 21

challenge
Show f(x)=x3βˆ’3x+1f(x) = x^3 - 3x + 1 has three real roots by locating sign changes.

Example 22

challenge
A continuous ff on [0,1][0,1] satisfies f(0)=f(1)f(0)=f(1). Show some cc has f(c)=f(c+12)f(c) = f(c + \tfrac12) on [0,12][0,\tfrac12].

Example 23

easy
A continuous function ff has f(0)=βˆ’4f(0)=-4 and f(3)=6f(3)=6. Does the IVT guarantee a value c∈(0,3)c \in (0,3) with f(c)=1f(c)=1?

Example 24

easy
Let f(x)=x2βˆ’5f(x)=x^2-5. Use IVT to show ff has a root in (2,3)(2,3).

Example 25

easy
A continuous function has f(βˆ’1)=2f(-1)=2 and f(2)=2f(2)=2. Does IVT guarantee f(c)=0f(c)=0 for some c∈(βˆ’1,2)c \in (-1,2)?

Example 26

easy
Can IVT be applied to f(x)=⌊xβŒ‹f(x)=\lfloor x \rfloor on [0.5,2.5][0.5,2.5]? Why or why not?

Example 27

medium
Show that f(x)=x4βˆ’3xβˆ’1f(x)=x^4-3x-1 has a root in (1,2)(1,2).

Example 28

medium
Use IVT to show ex=3βˆ’xe^x = 3-x has a solution in (0,1)(0,1).

Example 29

medium
A continuous function on [0,10][0,10] has f(0)=βˆ’7f(0)=-7 and f(10)=15f(10)=15. List three values IVT guarantees the function attains on (0,10)(0,10).

Example 30

medium
Show sin⁑x=x2\sin x = \frac{x}{2} has a positive solution.

Example 31

medium
At noon a car is at mile marker 5; at 1pm it is at mile marker 65. Assuming continuous motion, must it have passed mile marker 40 at some time in between?

Example 32

medium
Show that f(x)=x3βˆ’6x+2f(x)=x^3-6x+2 has at least two roots in [βˆ’3,3][-3,3].

Example 33

medium
A continuous function ff on [1,5][1,5] has f(1)=3f(1)=3 and f(5)=8f(5)=8. Must ff attain the value 1010 on the interval?

Example 34

hard
Show 2x=3x2^x = 3x has a solution in (0,1)(0,1) and another in (3,4)(3,4).

Example 35

hard
Let ff be continuous on [0,1][0,1] with f(0)=1f(0)=1 and f(1)=0f(1)=0. Prove there exists cc with f(c)=cf(c)=c.

Example 36

hard
Show that the polynomial p(x)=x5βˆ’4x3+xβˆ’1p(x)=x^5-4x^3+x-1 has at least three real roots.

Example 37

hard
Use IVT to approximate the root of f(x)=x3βˆ’xβˆ’2f(x)=x^3-x-2 in (1,2)(1,2) to one decimal place by bisection.

Example 38

hard
Two continuous functions f,g:[0,1]β†’Rf,g:[0,1] \to \mathbb{R} satisfy f(0)<g(0)f(0)<g(0) and f(1)>g(1)f(1)>g(1). Prove they intersect.

Example 39

challenge
Let ff be continuous on [0,2][0,2] with f(0)=f(2)f(0)=f(2). Prove there is c∈[0,1]c \in [0,1] with f(c)=f(c+1)f(c)=f(c+1).

Example 40

challenge
Show that the equation x7+x3βˆ’2xβˆ’5=0x^7 + x^3 - 2x - 5 = 0 has at least one real solution.

Example 41

challenge
A continuous f:[0,1]β†’Rf:[0,1] \to \mathbb{R} satisfies f(0)+f(1)=1f(0)+f(1)=1. Show some c∈[0,1]c \in [0,1] has f(c)=12f(c)=\tfrac12.
← Back to Intermediate Value Theorem Formula Explained Practice Mode

Background Knowledge

These ideas may be useful before you work through the harder examples.

limitcontinuity types