Unit 1 · Topic 1.16 Beta

The Intermediate Value Theorem

If f is continuous on [a, b] and d is between f(a) and f(b) , then f(c) = d for at least one c in (a, b) .

Practice 3: Justification

Part 1 · Hook

Why this matters

If you were 50 cm tall at birth and 170 cm tall at 16, then at some moment you were exactly 1 meter tall. Nobody measured it, but height does not jump, so it had to pass through every value in between. The Intermediate Value Theorem turns that common sense into a tool, and the exam rewards you for saying exactly why it applies.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which function is continuous on $[0, 2]$ ?

$x^2 - 3x$
$\frac{1}{x - 1}$
$\ln x$
$\frac{|x - 1|}{x - 1}$

Show the answer

A polynomial is continuous everywhere. The others break inside [0, 2] or at 0.

Correct: $x^2 - 3x$ :
$\frac{1}{x - 1}$ :
$\ln x$ :
$\frac{|x - 1|}{x - 1}$ :

2. At a jump discontinuity, a function

skips the values between its one-sided limits
grows without bound
has a hole
is undefined

Show the answer

The graph steps from one height to another with nothing in between.

Correct: skips the values between its one-sided limits:
grows without bound:
has a hole:
is undefined:

Part 4 · See it

See it first

Left: a continuous function on the closed interval [1, 5] with f(1) = 1 and f(5) = 5. The dashed line y = 3 lies between 1 and 5, and the curve crosses it three times, at c₁, c₂ and c₃, all inside (1, 5): the theorem promises at least one. Right: a function with a jump at x = 3 that also has f(1) = 1 and f(5) = 5 but skips over y = 3, so no c gives f(c) = 3. — Continuity on the closed interval forces every in-between value; a jump can skip one. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f$ is continuous on the closed interval $[a, b]$ its graph has no break from $(a, f(a))$ to $(b, f(b))$
A number $d$ lies between $f(a)$ and $f(b)$ the horizontal line $y = d$ separates the two endpoints of the graph
An unbroken graph joins points on opposite sides of a lineit must cross the line: $f(c) = d$ for at least one $c$ in $(a, b)$
Either hypothesis fails (a break, or $d$ outside the range of the endpoint values)the theorem promises nothing; the value may or may not be reached

Part 6 · Key ideas

Key ideas

Intermediate Value Theorem (IVT): if $f$ is continuous on $[a, b]$ and $d$ is between $f(a)$ and $f(b)$ , then $f(c) = d$ for at least one $c$ in $(a, b)$ .
Hypotheses come first: continuity on the closed interval, with a reason ("polynomial", "stated in the problem"), and $d$ between the endpoint values, with the numbers.
The conclusion is existence: at least one $c$ . The theorem does not say where, or how many.
With a table, apply the theorem on each pair of neighboring inputs; non-overlapping intervals give separate solutions.
Finding $c$ is a separate job: algebra, or a calculator (three decimal places).

Part 7 · Misconception

A common mistake

The wrong idea: $f(0) = -1$ and $f(4) = 3$ , so by the IVT, $f(c) = 1$ for some $c$ in $(0, 4)$ .

What actually happens: The theorem needs $f$ to be continuous on $[0, 4]$ , and the justification must say so. A function with a jump can have $f(0) < 1 < f(4)$ and never equal 1. Write: " $f$ is continuous on $[0, 4]$ and $f(0) < 1 < f(4)$ , so by the IVT ..."

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Temperature of a liquid

The temperature T(t) of a liquid, in degrees Celsius, at selected times t, in minutes. T is a continuous function of t.
$t$ (minutes)	0	2	5	7	10
$T(t)$ (degrees Celsius)	20	35	31	44	40

1. Must there be a time $t$ with $2 < t < 5$ at which $T(t) = 33$ ?

Yes: $T$ is continuous on $[2, 5]$ and $31 < 33 < 35$
No: $T$ decreases from 35 to 31 on that interval, so it skips 33
No: 33 is not one of the values in the table
Yes, at $t = 3.5$ , halfway between 2 and 5

Show the answer

$T$ is continuous on $[2, 5]$ , $T(2) = 35$ and $T(5) = 31$ , and 33 is between 31 and 35. By the Intermediate Value Theorem, $T(c) = 33$ for some $c$ in $(2, 5)$ .

Correct: Yes: $T$ is continuous on $[2, 5]$ and $31 < 33 < 35$ : Continuity on the closed interval plus a value between the endpoint values: the theorem applies.
No: $T$ decreases from 35 to 31 on that interval, so it skips 33: Decreasing or not, a continuous function passes through every value between its endpoint values.
No: 33 is not one of the values in the table: The table lists a few times only; the theorem is about all the times in between.
Yes, at $t = 3.5$ , halfway between 2 and 5: The theorem says such a time exists, not where it is.

2. What is the least number of times $t$ in $0 < t < 10$ at which $T(t) = 33$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Apply the theorem on each interval between table times. 33 is between 20 and 35 on $[0, 2]$ , between 35 and 31 on $[2, 5]$ , and between 31 and 44 on $[5, 7]$ , but not between 44 and 40 on $[7, 10]$ . The three open intervals do not overlap, so $T(t) = 33$ at least 3 times.

Answer: 3

3. Which is a complete justification that $T(c) = 30$ for some $c$ in $(0, 2)$ ?

$T(0) = 20$ and $T(2) = 35$ , and 30 is between 20 and 35, so by the IVT, $T(c) = 30$ for some $c$ in $(0, 2)$
$T$ is continuous on $[0, 2]$ , so by the Intermediate Value Theorem, $T(c) = 30$ for some value $c$ in $(0, 2)$
$T$ is continuous on $[0, 2]$ and $T(0)$ $= 20 < 30 < 35$ $= T(2)$ , so by the IVT, $T(c) = 30$ for some $c$ in $(0, 2)$
$T$ is continuous on $[0, 2]$ and $20 < 30 < 35$ , so by the Intermediate Value Theorem, $T(1) = 30$ at $t = 1$

Show the answer

A full justification states both conditions, continuity on the closed interval and the value between $T(0)$ and $T(2)$ , names the theorem, and concludes that some $c$ in the open interval works.

$T(0) = 20$ and $T(2) = 35$ , and 30 is between 20 and 35, so by the IVT, $T(c) = 30$ for some $c$ in $(0, 2)$ : It never says $T$ is continuous on $[0, 2]$ ; without that the theorem cannot be used.
$T$ is continuous on $[0, 2]$ , so by the Intermediate Value Theorem, $T(c) = 30$ for some value $c$ in $(0, 2)$ : It never shows that 30 is between $T(0)$ and $T(2)$ .
Correct: $T$ is continuous on $[0, 2]$ and $T(0)$ $= 20 < 30 < 35$ $= T(2)$ , so by the IVT, $T(c) = 30$ for some $c$ in $(0, 2)$ : Both conditions, the theorem's name, and the correct conclusion.
$T$ is continuous on $[0, 2]$ and $20 < 30 < 35$ , so by the Intermediate Value Theorem, $T(1) = 30$ at $t = 1$ : The theorem gives some $c$ , not a particular one such as $t = 1$ .

Graph

The graph of $g$ on $[0, 4]$

4. $g(0) = -1$ and $g(4) = 3$ . Does the Intermediate Value Theorem guarantee some $c$ in $(0, 4)$ with $g(c) = 1$ ?

Yes: 1 is between $g(0) = -1$ and $g(4) = 3$
Yes: $g$ is defined at each single point of the interval $[0, 4]$
No: 1 is not between $g(0) = -1$ and $g(4) = 3$
No: $g$ is not continuous on $[0, 4]$ , since it jumps at $x = 2$

Show the answer

The theorem needs $g$ continuous on $[0, 4]$ . At $x = 2$ the one-sided limits are 0.5 and 2, so $g$ is not continuous there, and the theorem does not apply.

Yes: 1 is between $g(0) = -1$ and $g(4) = 3$ : Being between the endpoint values is only one of the two hypotheses.
Yes: $g$ is defined at each single point of the interval $[0, 4]$ : Defined is not the same as continuous.
No: 1 is not between $g(0) = -1$ and $g(4) = 3$ : 1 is between $-1$ and 3; the problem is continuity.
Correct: No: $g$ is not continuous on $[0, 4]$ , since it jumps at $x = 2$ : The continuity hypothesis fails at the jump, so no guarantee.

5. Let $f(x) = x^3 - 4x + 1$ . Then $f(0) = 1$ and $f(1) = -2$ . Which statement justifies that $f$ has a zero in $(0, 1)$ ?

$f(0) = 1$ is positive and $f(1) = -2$ is negative, so the graph of $f$ crosses the $x$ -axis at $x = 0.5$
$f$ is a polynomial, so it has a zero; since $f(0)$ and $f(1)$ differ, that zero lies between 0 and 1
$f$ is continuous on $[0, 1]$ , so by the Intermediate Value Theorem it equals 0 somewhere in $(0, 1)$
$f$ is a polynomial, so it is continuous on $[0, 1]$ ; since $-2 < 0 < 1$ , the IVT gives $f(c) = 0$ for some $c$ in $(0, 1)$

Show the answer

Name why $f$ is continuous on the closed interval (polynomial), show 0 is between $f(0)$ and $f(1)$ , name the theorem, and conclude with some $c$ in $(0, 1)$ .

$f(0) = 1$ is positive and $f(1) = -2$ is negative, so the graph of $f$ crosses the $x$ -axis at $x = 0.5$ : It never mentions continuity, and the theorem does not give the location 0.5.
$f$ is a polynomial, so it has a zero; since $f(0)$ and $f(1)$ differ, that zero lies between 0 and 1: Not every polynomial has a real zero, and "differ" is not enough; 0 must lie between the values.
$f$ is continuous on $[0, 1]$ , so by the Intermediate Value Theorem it equals 0 somewhere in $(0, 1)$ : It never shows that 0 is between $f(0) = 1$ and $f(1) = -2$ .
Correct: $f$ is a polynomial, so it is continuous on $[0, 1]$ ; since $-2 < 0 < 1$ , the IVT gives $f(c) = 0$ for some $c$ in $(0, 1)$ : Reason for continuity, the value between the endpoint values, the theorem, and the conclusion.

6. $f(x) = \frac{1}{x}$ has $f(-1) = -1$ and $f(1) = 1$ , yet $f(x) = 0$ has no solution. Why does this not contradict the Intermediate Value Theorem?

0 is not between $f(-1) = -1$ and $f(1) = 1$
$f$ is not continuous on $[-1, 1]$ ; it is undefined at $x = 0$
The theorem applies to polynomials, and $\frac{1}{x}$ is not one
The theorem needs $f(-1)$ and $f(1)$ to have the same sign

Show the answer

The theorem's hypothesis is continuity on the closed interval $[-1, 1]$ . $\frac{1}{x}$ has an infinite discontinuity at 0, so the hypothesis fails and the theorem promises nothing.

0 is not between $f(-1) = -1$ and $f(1) = 1$ : 0 is between $-1$ and $1$ ; that hypothesis holds.
Correct: $f$ is not continuous on $[-1, 1]$ ; it is undefined at $x = 0$ : A failed hypothesis means no conclusion, so no contradiction.
The theorem applies to polynomials, and $\frac{1}{x}$ is not one: The theorem applies to every function continuous on $[a, b]$ , polynomial or not.
The theorem needs $f(-1)$ and $f(1)$ to have the same sign: Opposite signs are what put 0 between them.

Part 9 · Summary

Summary

f

is continuous on

[a, b]

and

d

is between

f(a)

and

f(b)

, then

f(c) = d

for at least one

c

(a, b)

. A justification states both hypotheses with reasons and numbers, names the theorem, and concludes "for some

c

(a, b)

". Without continuity on the closed interval, the conclusion can fail.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Temperature of a liquid

The graph of gg on [0,4][0, 4]

Summary

What comes next

Connections

The graph of $g$ on $[0, 4]$