Intermediate Value Theorem: Worked examples — Maths deep dive

Before anything, one reminder of the meaning of each symbol we'll reuse:

The scenario matrix

Every problem IVT can throw at you falls into one of these cells. The last column names the example that clears it.

Cell	Situation	What makes it tricky	Example
A	$f (a) < 0 < f (b)$ (opposite signs)	the classic Bolzano root	Ex 1
B	$f (a) > 0 > f (b)$ (opposite signs, flipped)	sign order reversed	Ex 2
C	Hit a specific $N \neq = 0$	must build $g = f - N$ first	Ex 3
D	Same sign at both ends	IVT is silent — trap	Ex 4
E	Endpoint exactly equals $N$	is $c$ allowed to be $a$ or $b$ ?	Ex 5
F	Discontinuous input	hypothesis fails — no guarantee	Ex 6
G	Degenerate: $f (a) = f (b)$	no strictly-between $N$ exists	Ex 7
H	Fixed point / self-referential	turn $f (x) = x$ into a root	Ex 8
I	Real-world word problem	model, then apply	Ex 9
J	Exam twist: count / locate	existence vs uniqueness	Ex 10
K	$N$ outside $[f (a), f (b)]$	IVT silent (not the same as D)	Ex 11

Skim it, then work down. Each example labels its cell.

Cell A — opposite signs, the classic root

Worked example Example 1 (Cell A)

Show $f (x) = x^{3} - 4 x + 1$ has a root in $[0, 1]$ .

Forecast: guess now — does a root live in $[0, 1]$ ? Yes / No?

$f$ is a polynomial $\Rightarrow$ continuous everywhere. Why this step? IVT's price of admission is Continuity on the closed interval. Polynomials always pay it, so we're allowed to proceed.
Evaluate the ends: $f (0) = 0 - 0 + 1 = 1$ and $f (1) = 1 - 4 + 1 = - 2$ . Why this step? The two end-heights frame the range of values the curve is forced to sweep through.
The signs are opposite ( $+ 1$ then $- 2$ ), so $0$ lies strictly between $- 2$ and $1$ . Why this step? $N = 0$ being an intermediate value is exactly the condition IVT needs.
By IVT, $\exists c \in (0, 1)$ with $f (c) = 0$ . ✓

Verify: the sign product $f (0) \cdot f (1) = (1) (- 2) = - 2 < 0$ . A negative product means the ends straddle zero — the smell test passes.

Look at the figure: the white curve dives from height $+ 1$ down to $- 2$ , and the amber crossing point is the $c$ IVT promises.

Cell B — opposite signs, flipped order

The figure shows the two curves $y = cos x$ and $y = x$ crossing, and equivalently $f = cos x - x$ cutting the axis.

Cell C — hit a specific value $N \neq = 0$

The figure shows $f (x) = x^{2} + x$ climbing across the horizontal target line $y = 5$ .

Cell D — same sign at both ends (the silent trap)

Worked example Example 4 (Cell D)

A student writes: " $f (x) = x^{2} - 2$ has $f (- 2) = 2 > 0$ and $f (2) = 2 > 0$ , same sign, so no root in $[- 2, 2]$ ." Is this correct?

Forecast: is the student right or wrong?

$f (- 2) = 4 - 2 = 2$ and $f (2) = 4 - 2 = 2$ . Both positive — same sign. Why this step? We confirm the student's arithmetic; it's fine.
IVT with $N = 0$ : is $0$ between $f (- 2) = 2$ and $f (2) = 2$ ? No — both ends are $2$ , so $0$ is not intermediate. Why this step? IVT's conclusion simply does not fire. It says nothing.
But "IVT is silent" is not "no root exists". Check directly: $x^{2} - 2 = 0 \Rightarrow x = \pm 2 \approx \pm 1.414$ , both inside $[- 2, 2]$ . Why this step? IVT is a sufficient tool, not a necessary one. Same signs = no information, so we must look ourselves.
Conclusion: the student's logic is wrong. There are in fact two roots. ✗ to the student.

Verify: $(2)^{2} - 2 = 0$ and $(- 2)^{2} - 2 = 0$ . Two roots, both in range, despite matching endpoint signs.

The figure shows the parabola dipping below the axis between two equal, positive endpoints — the dip IVT cannot see.

Cell E — endpoint exactly equals the target

Worked example Example 5 (Cell E)

Where does $f (x) = x^{2}$ equal $N = 4$ on $[0, 2]$ , and is the promised $c$ strictly inside $(0, 2)$ ?

Forecast: will $c$ be an endpoint or an interior point?

$f (0) = 0$ , $f (2) = 4$ . Is $N = 4$ strictly between $0$ and $4$ ? No — $4$ equals the right end $f (2)$ . Why this step? IVT requires $N$ strictly between $f (a)$ and $f (b)$ to guarantee an interior $c \in (a, b)$ . Here $N$ sits on the boundary.
So IVT gives no interior guarantee. But the value is still attained — at the endpoint $x = 2$ itself, since $f (2) = 4$ . Why this step? The value being hit and being hit inside the open interval are different claims. IVT's strict-between clause is exactly what forbids the endpoint case.
Is there also an interior solution? $x^{2} = 4 \Rightarrow x = \pm 2$ ; only $x = 2$ is in $[0, 2]$ , and it's the endpoint. Why this step? We check no other input works. It doesn't — the only hit is the boundary.
Lesson: when $N$ equals an endpoint height, the " $c \in (a, b)$ " (open) conclusion can fail even though the value is reached.

Verify: $f (2) = 2^{2} = 4 = N$ . Attained, but at $c = 2 \in / (0, 2)$ . This is why the theorem demands $N$ strictly between.

Cell F — discontinuous input, hypothesis fails

Worked example Example 6 (Cell F)

Does $f (x) = {- 1 1 x < 0 x \geq 0$ hit the value $N = 0$ on $[- 1, 1]$ ?

Forecast: the curve goes from $- 1$ up to $+ 1$ — must it pass through $0$ ?

Ends: $f (- 1) = - 1 < 0$ and $f (1) = 1 > 0$ . Opposite signs — looks like a slam-dunk for IVT. Why this step? We deliberately set up the appearance of Cell A to expose the trap.
Check the hypothesis: is $f$ continuous on $[- 1, 1]$ ? No — at $x = 0$ it jumps from $- 1$ straight to $+ 1$ with nothing in between. Why this step? IVT's engine is Continuity. No continuity, no promise. This is the step everyone skips.
Since the hypothesis fails, IVT says nothing. Look directly: $f$ only ever outputs $- 1$ or $+ 1$ ; it never equals $0$ . Why this step? A jump lets the function teleport over $N = 0$ — exactly the "skip" continuity forbids.
Answer: no, $0$ is never attained. Opposite signs alone are not enough. ✗

Verify: the range of $f$ is ${- 1, 1}$ ; $0 \in / {- 1, 1}$ , so $f (x) = 0$ has no solution. Mirrors the $1/ x$ mistake in the parent note.

The figure shows the jump discontinuity leaping over $N = 0$ (dashed line) — no crossing.

Cell G — degenerate: equal endpoints

Worked example Example 7 (Cell G)

On $[0, 2 π]$ , take $f (x) = sin x$ . What does IVT guarantee about a value $N$ strictly between $f (0)$ and $f (2 π)$ ?

Forecast: how many intermediate values $N$ does IVT hand you here?

$f (0) = sin 0 = 0$ and $f (2 π) = sin 2 π = 0$ . The two endpoint heights are equal. Why this step? The parent's definition requires $f (a) \neq = f (b)$ to have any $N$ strictly between them.
If $f (a) = f (b) = 0$ , the set of numbers "strictly between $0$ and $0$ " is empty. Why this step? There is no room between a number and itself, so IVT's premise has no valid $N$ to offer.
So IVT is vacuous here — it names no guaranteed value. Why this step? Vacuous ≠ false. It just means this theorem is the wrong tool for equal endpoints. (For "must the slope be zero somewhere?" you'd reach for Rolle's Theorem instead.)
Directly, of course $sin$ does hit values in $[- 1, 1]$ on the way — but that's from interior behaviour, not from comparing the two equal ends.

Verify: $sin (0) = 0$ and $sin (2 π) = 0$ are equal, so no $N$ satisfies $min < N < max$ of the endpoints. The strictly-between set is empty. ✓ (vacuous case).

Cell H — fixed point (self-referential)

Cell I — real-world word problem

Worked example Example 9 (Cell I)

A ball is thrown up so its height (in metres) is $h (t) = 1 + 14 t - 5 t^{2}$ for time $t \in [0, 3]$ seconds. Show there is an instant in $(0, 3)$ when the ball is exactly $8$ m high.

Forecast: guess now — does the ball reach exactly $8$ m at some instant strictly inside the flight?

$h$ is a polynomial in $t$ $\Rightarrow$ continuous; a thrown ball changes height smoothly (no teleporting). Why this step? Physical motion is continuous — IVT's hypothesis is naturally satisfied.
Left end: $h (0) = 1 + 0 - 0 = 1$ m, so $h (0) = 1 < 8$ . Why this step? The ball starts below the target height $N = 8$ .
Right end: $h (3) = 1 + 42 - 45 = - 2$ m, so $h (3) = - 2 < 8$ ... but this is also below $8$ — same side. So the plain endpoints do not straddle $8$ . Why this step? We must not fake an opposite-sign setup; the ball has already fallen back down by $t = 3$ . We need a smarter interval.
Pick an interior probe at the top of the arc. Try $t = 1$ : $h (1) = 1 + 14 - 5 = 10$ m, so $h (1) = 10 > 8$ . Why this step? We hunt for one point above $N = 8$ to pair with a point below it, so that $8$ becomes an intermediate value on a sub-interval.
Now apply IVT on $[0, 1]$ : $h$ continuous, $h (0) = 1 < 8$ and $h (1) = 10 > 8$ , so $8$ is strictly between $1$ and $10$ . IVT $\Rightarrow \exists c_{1} \in (0, 1)$ with $h (c_{1}) = 8$ . ✓ Why this step? This is a genuine IVT existence argument — the endpoints of $[0, 1]$ truly straddle $8$ , unlike the full interval.
Bonus (coming down): apply IVT on $[1, 3]$ : $h (1) = 10 > 8$ and $h (3) = - 2 < 8$ , so IVT gives a second instant $c_{2} \in (1, 3)$ with $h (c_{2}) = 8$ . ✓ Why this step? The ball rises through $8$ m and later falls back through it — two crossings, both found by IVT, neither computed by hand.

Verify: solve $1 + 14 t - 5 t^{2} = 8 \Rightarrow 5 t^{2} - 14 t + 7 = 0 \Rightarrow t = \frac{14 \pm 196 - 140}{10} = \frac{14 \pm 56}{10}$ . So $t \approx 0.6517$ (in $(0, 1)$ ) and $t \approx 2.148$ (in $(1, 3)$ ). Units: seconds; both instants lie inside $(0, 3)$ . Two genuine intermediate-value crossings that IVT predicted without solving.

Cell J — exam twist: existence vs count

Worked example Example 10 (Cell J)

Exam question: "Using IVT, how many solutions does $x^{3} - 3 x = 0$ have on $[- 2, 2]$ ?" A student answers "IVT says at least one." Grade this and give the true count.

Forecast: how many roots, and is "at least one" the best you can say?

$f (x) = x^{3} - 3 x$ , continuous. $f (- 2) = - 8 + 6 = - 2 < 0$ , $f (2) = 8 - 6 = 2 > 0$ : opposite signs. Why this step? IVT with $N = 0$ fires — guarantees at least one root in $(- 2, 2)$ . The student's IVT claim is correct as far as IVT goes.
But IVT is pure existence — it never counts. Saying "exactly one" from IVT alone is the trap. Why this step? The parent's mistake list: IVT gives no count and no formula for $c$ .
Find the true count by factoring: $x^{3} - 3 x = x (x^{2} - 3) = x (x - 3) (x + 3)$ . Roots at $x = 0, 3 \approx 1.732, - 3 \approx - 1.732$ — all three lie in $[- 2, 2]$ . Why this step? To count roots we need algebra/monotonicity, not IVT.
Verdict: student is partially right (existence yes) but IVT alone cannot give the count. True count is three. For uniqueness/count you'd use monotonicity or Rolle's Theorem.

Verify: $f (0) = 0$ , $f (3) = 33 - 33 = 0$ , $f (- 3) = 0$ . Three roots in $[- 2, 2]$ ; IVT only promised $\geq 1$ .

Cell K — target outside the endpoint range

Worked example Example 11 (Cell K)

Does IVT guarantee $f (x) = x^{2}$ hits $N = 9$ somewhere on $[0, 2]$ ?

Forecast: is $9$ reachable on this interval, and can IVT tell you?

Endpoints: $f (0) = 0$ and $f (2) = 4$ . The endpoint range is $[0, 4]$ . Why this step? IVT only speaks about $N$ between $f (a)$ and $f (b)$ — here between $0$ and $4$ .
Is $N = 9$ between $0$ and $4$ ? No — $9 > 4$ , so $N$ lies outside $[0, 4]$ , above both endpoints. Why this step? This is different from Cell D (same sign): there the endpoints matched; here $N$ simply overshoots the whole endpoint band.
IVT's premise " $N$ intermediate" fails, so IVT is silent — it neither promises nor forbids a solution. Why this step? Silence again means "wrong tool", not "no answer".
Direct check: on $[0, 2]$ the largest value of $x^{2}$ is $f (2) = 4 < 9$ , so $9$ is genuinely never reached here. (Widen to $[0, 3]$ and $f (3) = 9$ would appear.) ✗ on $[0, 2]$ . Why this step? We confirm by hand what IVT declined to rule on — the maximum output on the interval is $4$ , below $9$ .

Verify: $max_{[0, 2]} x^{2} = 4$ and $9 > 4$ , so $x^{2} = 9$ has no solution in $[0, 2]$ ; IVT was correctly silent, and this is not a same-sign trap.

Recap of the matrix

Connections

Intermediate Value Theorem — the parent rule these examples exercise.
Continuity — Cell F fails precisely because this breaks.
Bolzano's Theorem — the $N = 0$ engine behind Cells A, B, H, J.
Bisection Method — how to actually locate the $c$ IVT promises.
Rolle's Theorem and Mean Value Theorem — the tools for counting/uniqueness (Cell J).
Fixed Point Theorems — Cell H's natural home.