Intermediate Value Theorem: Worked examples — Hinglish deep dive

4.1.8 · D3 · Maths › Calculus I — Limits & Derivatives › Intermediate Value Theorem

Kuch bhi shuru karne se pehle, ek reminder ki har woh symbol kya matlab rakhta hai jo hum reuse karenge:

Scenario matrix

IVT ka har problem in cells mein se ek mein fit hota hai. Last column us example ka naam deta hai jo use clear karta hai.

Cell	Situation	Kya mushkil hai	Example
A	$f (a) < 0 < f (b)$ (opposite signs)	classic Bolzano root	Ex 1
B	$f (a) > 0 > f (b)$ (opposite signs, flipped)	sign order ulta hai	Ex 2
C	Ek specific $N \neq = 0$ hit karo	pehle $g = f - N$ banana padega	Ex 3
D	Dono ends par same sign	IVT chup hai — trap	Ex 4
E	Endpoint exactly $N$ ke barabar	kya $c$ , $a$ ya $b$ ho sakta hai?	Ex 5
F	Discontinuous input	hypothesis fail — koi guarantee nahi	Ex 6
G	Degenerate: $f (a) = f (b)$	koi strictly-between $N$ exist nahi karta	Ex 7
H	Fixed point / self-referential	$f (x) = x$ ko root mein badlo	Ex 8
I	Real-world word problem	model karo, phir apply karo	Ex 9
J	Exam twist: count / locate	existence vs uniqueness	Ex 10
K	$N$ bahar $[f (a), f (b)]$ ke	IVT chup (D jaisa nahi)	Ex 11

Pehle ek nazar dalo, phir neeche kaam karo. Har example apna cell label karta hai.

Cell A — opposite signs, classic root

Worked example Example 1 (Cell A)

Dikhao ki $f (x) = x^{3} - 4 x + 1$ ka $[0, 1]$ mein ek root hai.

Forecast: abhi andaza lagao — kya $[0, 1]$ mein koi root rehta hai? Haan / Nahi?

$f$ ek polynomial hai $\Rightarrow$ har jagah continuous hai. Yeh step kyun? IVT ka admission price closed interval par Continuity hai. Polynomials hamesha yeh pay karte hain, toh hum aage badh sakte hain.
Ends evaluate karo: $f (0) = 0 - 0 + 1 = 1$ aur $f (1) = 1 - 4 + 1 = - 2$ . Yeh step kyun? Dono end-heights woh range frame karti hain jis se hokar curve ko zaroor guzarna padta hai.
Signs opposite hain ( $+ 1$ phir $- 2$ ), toh $0$ , $- 2$ aur $1$ ke beech strictly baithe hai. Yeh step kyun? $N = 0$ ka intermediate value hona exactly wahi condition hai jo IVT ko chahiye.
IVT se, $\exists c \in (0, 1)$ with $f (c) = 0$ . ✓

Verify: sign product $f (0) \cdot f (1) = (1) (- 2) = - 2 < 0$ . Negative product matlab ends zero ko straddle kar rahe hain — smell test pass.

Figure dekho: white curve height $+ 1$ se $- 2$ tak neeche jaati hai, aur amber crossing point woh $c$ hai jo IVT promise karta hai.

Cell B — opposite signs, flipped order

Worked example Example 2 (Cell B)

Dikhao ki $cos x = x$ ka $[0, 1]$ mein ek solution hai.

Forecast: kya cosine curve kabhi $0$ aur $1$ ke beech line $y = x$ ke barabar hoti hai?

Maano $f (x) = cos x - x$ . Yeh continuous hai (cosine aur $x$ dono hain, aur unka difference bhi). Yeh step kyun? Hum " $cos x = x$ " ko " $f (x) = 0$ " mein rewrite karte hain taaki yeh root hunt ban jaaye — parent se Move 2.
$f (0) = cos 0 - 0 = 1 > 0$ . Yeh step kyun? Left end positive hai.
$f (1) = cos 1 - 1 \approx 0.5403 - 1 = - 0.4597 < 0$ . Yeh step kyun? Right end negative hai — note karo ki sign order $+$ phir $-$ hai, Ex 1 ka ulta. IVT ko order ki parwah nahi, bas yeh ki woh differ karein.
Opposite signs $\Rightarrow$ IVT $c \in (0, 1)$ deta hai with $cos c = c$ . ✓

Verify: product $f (0) \cdot f (1) \approx (1) (- 0.4597) = - 0.4597 < 0$ . Zero ko straddle kar raha hai. (Actual root $c \approx 0.739$ ke paas hai, famous "Dottie number".)

Figure mein do curves $y = cos x$ aur $y = x$ cross karti hain, aur equivalently $f = cos x - x$ axis ko cut karta hai.

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

Figure mein $f (x) = x^{2} + x$ horizontal target line $y = 5$ par chardhta dikhta hai.

Cell D — dono ends par same sign (silent trap)

Worked example Example 4 (Cell D)

Ek student likhta hai: " $f (x) = x^{2} - 2$ mein $f (- 2) = 2 > 0$ aur $f (2) = 2 > 0$ hain, same sign, toh $[- 2, 2]$ mein koi root nahi." Kya yeh sahi hai?

Forecast: kya student sahi hai ya galat?

$f (- 2) = 4 - 2 = 2$ aur $f (2) = 4 - 2 = 2$ . Dono positive — same sign. Yeh step kyun? Hum student ki arithmetic confirm karte hain; woh theek hai.
IVT with $N = 0$ : kya $0$ , $f (- 2) = 2$ aur $f (2) = 2$ ke beech hai? Nahi — dono ends $2$ hain, toh $0$ intermediate nahi hai. Yeh step kyun? IVT ka conclusion bas fire nahi hota. Woh kuch nahi kehta.
Lekin "IVT chup hai" ka matlab "koi root exist nahi karta" nahi hai. Seedha check karo: $x^{2} - 2 = 0 \Rightarrow x = \pm 2 \approx \pm 1.414$ , dono $[- 2, 2]$ ke andar. Yeh step kyun? IVT ek sufficient tool hai, necessary nahi. Same signs = koi information nahi, toh hum khud dekhte hain.
Conclusion: student ki logic galat hai. Wakai mein do roots hain. ✗ student ke liye.

Verify: $(2)^{2} - 2 = 0$ aur $(- 2)^{2} - 2 = 0$ . Do roots, dono range mein, matching endpoint signs ke bawajood.

Figure mein parabola do equal, positive endpoints ke beech axis ke neeche dip karti hai — woh dip jo IVT dekh nahi sakta.

Cell E — endpoint exactly target ke barabar

Worked example Example 5 (Cell E)

$f (x) = x^{2}$ , $N = 4$ ko $[0, 2]$ par kahan hit karta hai, aur kya promised $c$ strictly $(0, 2)$ ke andar hai?

Forecast: kya $c$ endpoint hoga ya interior point?

$f (0) = 0$ , $f (2) = 4$ . Kya $N = 4$ , $0$ aur $4$ ke strictly beech hai? Nahi — $4$ , right end $f (2)$ ke barabar hai. Yeh step kyun? IVT ko $N$ strictly $f (a)$ aur $f (b)$ ke beech chahiye taaki interior $c \in (a, b)$ guarantee ho sake. Yahan $N$ boundary par baitha hai.
Toh IVT koi interior guarantee nahi deta. Lekin value phir bhi attain hoti hai — endpoint $x = 2$ par hi, kyunki $f (2) = 4$ . Yeh step kyun? Value ka hit hona aur open interval ke andar hit hona alag claims hain. IVT ka strict-between clause exactly wahi hai jo endpoint case ko rokata hai.
Kya koi interior solution bhi hai? $x^{2} = 4 \Rightarrow x = \pm 2$ ; sirf $x = 2$ hi $[0, 2]$ mein hai, aur woh endpoint hai. Yeh step kyun? Hum check karte hain ki koi aur input kaam karta hai ya nahi. Nahi karta — sirf boundary par hi hit hoti hai.
Lesson: jab $N$ ek endpoint height ke barabar ho, toh " $c \in (a, b)$ " (open) ka conclusion fail ho sakta hai chahe value reach ho jaaye.

Verify: $f (2) = 2^{2} = 4 = N$ . Attain hua, lekin $c = 2 \in / (0, 2)$ par. Isi liye theorem $N$ ko strictly beech mein demand karta hai.

Cell F — discontinuous input, hypothesis fail

Worked example Example 6 (Cell F)

Kya $f (x) = {- 1 1 x < 0 x \geq 0$ , $[- 1, 1]$ par $N = 0$ value hit karta hai?

Forecast: curve $- 1$ se $+ 1$ tak jaati hai — kya use $0$ se zaroor guzarna padega?

Ends: $f (- 1) = - 1 < 0$ aur $f (1) = 1 > 0$ . Opposite signs — lagta hai IVT ke liye slam-dunk hai. Yeh step kyun? Hum deliberately Cell A ka appearance set karte hain trap expose karne ke liye.
Hypothesis check karo: kya $f$ , $[- 1, 1]$ par continuous hai? Nahi — $x = 0$ par yeh $- 1$ se seedha $+ 1$ tak jump karta hai, beech mein kuch nahi. Yeh step kyun? IVT ka engine Continuity hai. Continuity nahi, promise nahi. Yahi step hai jo sablog skip karte hain.
Hypothesis fail hone ki wajah se, IVT kuch nahi kehta. Seedha dekho: $f$ sirf $- 1$ ya $+ 1$ output karta hai; yeh kabhi $0$ ke barabar nahi hota. Yeh step kyun? Ek jump function ko $N = 0$ ke upar teleport karne deta hai — exactly woh "skip" jo continuity forbid karti hai.
Jawab: nahi, $0$ kabhi attain nahi hota. Sirf opposite signs kaafi nahi hain. ✗

Verify: $f$ ki range ${- 1, 1}$ hai; $0 \in / {- 1, 1}$ , toh $f (x) = 0$ ka koi solution nahi. Parent note mein $1/ x$ mistake ko mirror karta hai.

Figure mein jump discontinuity $N = 0$ (dashed line) ke upar se leapti hai — koi crossing nahi.

Cell G — degenerate: equal endpoints

Worked example Example 7 (Cell G)

$[0, 2 π]$ par, $f (x) = sin x$ lo. IVT kisi value $N$ ke baare mein kya guarantee deta hai jo $f (0)$ aur $f (2 π)$ ke strictly beech ho?

Forecast: IVT yahan kitne intermediate values $N$ deta hai?

$f (0) = sin 0 = 0$ aur $f (2 π) = sin 2 π = 0$ . Dono endpoint heights barabar hain. Yeh step kyun? Parent ki definition ko koi $N$ strictly beech rakhne ke liye $f (a) \neq = f (b)$ chahiye.
Agar $f (a) = f (b) = 0$ , toh " $0$ aur $0$ ke strictly beech" wale numbers ka set empty hai. Yeh step kyun? Ek number aur khud ke beech mein koi jagah nahi hoti, toh IVT ki premise ka koi valid $N$ nahi hota.
Toh IVT yahan vacuous hai — woh koi guaranteed value nahi deta. Yeh step kyun? Vacuous ≠ false. Matlab bas yeh hai ki yeh theorem equal endpoints ke liye galat tool hai. ("Kya slope kahin zero hoga?" ke liye tum Rolle's Theorem use karte.)
Directly, bilkul $sin$ raste mein $[- 1, 1]$ mein values hit karta hai — lekin woh interior behaviour se hai, dono equal ends compare karne se nahi.

Verify: $sin (0) = 0$ aur $sin (2 π) = 0$ barabar hain, toh koi $N$ , $min < N < max$ of the endpoints satisfy nahi karta. Strictly-between set empty hai. ✓ (vacuous case).

Cell H — fixed point (self-referential)

Cell I — real-world word problem

Worked example Example 9 (Cell I)

Ek ball upar phenki jaati hai toh uski height (metres mein) $h (t) = 1 + 14 t - 5 t^{2}$ hai time $t \in [0, 3]$ seconds ke liye. Dikhao ki $(0, 3)$ mein koi ek instant hai jab ball exactly $8$ m uunchi hai.

Forecast: abhi andaza lagao — kya ball flight ke strictly andar kisi instant par exactly $8$ m reach karti hai?

$h$ , $t$ mein ek polynomial hai $\Rightarrow$ continuous; ek phenki gayi ball smoothly height change karti hai (koi teleporting nahi). Yeh step kyun? Physical motion continuous hoti hai — IVT ki hypothesis naturally satisfied hai.
Left end: $h (0) = 1 + 0 - 0 = 1$ m, toh $h (0) = 1 < 8$ . Yeh step kyun? Ball target height $N = 8$ se neeche shuru hoti hai.
Right end: $h (3) = 1 + 42 - 45 = - 2$ m, toh $h (3) = - 2 < 8$ ... lekin yeh bhi $8$ se neeche hai — same side. Toh plain endpoints $8$ ko straddle nahi karte. Yeh step kyun? Hume fake opposite-sign setup nahi banana chahiye; ball $t = 3$ tak wapas neeche aa chuki hai. Hume smarter interval chahiye.
Arc ke top par ek interior probe lo. $t = 1$ try karo: $h (1) = 1 + 14 - 5 = 10$ m, toh $h (1) = 10 > 8$ . Yeh step kyun? Hum ek point upar $N = 8$ ke dhoondh rahe hain jo kisi neeche wale point ke saath pair ho sake, taaki $8$ ek sub-interval par intermediate value ban jaaye.
Ab $[0, 1]$ par IVT apply karo: $h$ continuous, $h (0) = 1 < 8$ aur $h (1) = 10 > 8$ , toh $8$ strictly $1$ aur $10$ ke beech hai. IVT $\Rightarrow \exists c_{1} \in (0, 1)$ with $h (c_{1}) = 8$ . ✓ Yeh step kyun? Yeh genuine IVT existence argument hai — $[0, 1]$ ke endpoints sach mein $8$ ko straddle karte hain, poore interval ke unlike.
Bonus (neeche aate waqt): $[1, 3]$ par IVT apply karo: $h (1) = 10 > 8$ aur $h (3) = - 2 < 8$ , toh IVT ek doosra instant $c_{2} \in (1, 3)$ deta hai with $h (c_{2}) = 8$ . ✓ Yeh step kyun? Ball $8$ m se upar jaati hai aur baad mein wapas $8$ m se guzarti hai — do crossings, dono IVT se mile, koi haath se calculate nahi hua.

Verify: $1 + 14 t - 5 t^{2} = 8 \Rightarrow 5 t^{2} - 14 t + 7 = 0 \Rightarrow t = \frac{14 \pm 196 - 140}{10} = \frac{14 \pm 56}{10}$ solve karo. Toh $t \approx 0.6517$ ( $(0, 1)$ mein) aur $t \approx 2.148$ ( $(1, 3)$ mein). Units: seconds; dono instants $(0, 3)$ ke andar. Do genuine intermediate-value crossings jo IVT ne solve kiye bina predict ki.

Cell J — exam twist: existence vs count

Worked example Example 10 (Cell J)

Exam question: "IVT use karke, $[- 2, 2]$ par $x^{3} - 3 x = 0$ ke kitne solutions hain?" Ek student jawab deta hai "IVT kehta hai kam se kam ek." Ise grade karo aur sahi count do.

Forecast: kitne roots hain, aur kya "at least one" kehna best hai?

$f (x) = x^{3} - 3 x$ , continuous. $f (- 2) = - 8 + 6 = - 2 < 0$ , $f (2) = 8 - 6 = 2 > 0$ : opposite signs. Yeh step kyun? $N = 0$ ke saath IVT fire hota hai — $(- 2, 2)$ mein kam se kam ek root guarantee karta hai. Student ka IVT claim, jahan tak IVT ki baat hai, sahi hai.
Lekin IVT pure existence hai — yeh kabhi count nahi karta. Sirf IVT se "exactly one" kehna trap hai. Yeh step kyun? Parent ki mistake list: IVT koi count nahi deta aur $c$ ka koi formula nahi.
Factoring se sahi count nikalo: $x^{3} - 3 x = x (x^{2} - 3) = x (x - 3) (x + 3)$ . Roots $x = 0, 3 \approx 1.732, - 3 \approx - 1.732$ par — teeno $[- 2, 2]$ mein hain. Yeh step kyun? Roots count karne ke liye algebra/monotonicity chahiye, IVT nahi.
Verdict: student partially sahi hai (existence haan) lekin sirf IVT count nahi de sakta. Sahi count teen hai. Uniqueness/count ke liye monotonicity ya Rolle's Theorem use karo.

Verify: $f (0) = 0$ , $f (3) = 33 - 33 = 0$ , $f (- 3) = 0$ . $[- 2, 2]$ mein teen roots; IVT ne sirf $\geq 1$ promise kiya tha.

Cell K — target endpoint range ke bahar

Worked example Example 11 (Cell K)

Kya IVT guarantee karta hai ki $f (x) = x^{2}$ , $[0, 2]$ par kahin $N = 9$ hit karega?

Forecast: kya $9$ is interval par reachable hai, aur kya IVT bata sakta hai?

Endpoints: $f (0) = 0$ aur $f (2) = 4$ . Endpoint range $[0, 4]$ hai. Yeh step kyun? IVT sirf un $N$ ke baare mein bolta hai jo $f (a)$ aur $f (b)$ ke beech hain — yahan $0$ aur $4$ ke beech.
Kya $N = 9$ , $0$ aur $4$ ke beech hai? Nahi — $9 > 4$ , toh $N$ poore endpoint band se upar hai. Yeh step kyun? Yeh Cell D se alag hai (same sign): wahan endpoints match karte the; yahan $N$ sirf poore endpoint band ko overshoot karta hai.
IVT ki premise " $N$ intermediate" fail hoti hai, toh IVT chup hai — woh na promise karta hai na forbid. Yeh step kyun? Yahan bhi silence ka matlab "galat tool" hai, "koi jawab nahi" nahi.
Direct check: $[0, 2]$ par $x^{2}$ ki sabse badi value $f (2) = 4 < 9$ hai, toh $9$ genuinely kabhi reach nahi hoti. ( $[0, 3]$ tak widen karo aur $f (3) = 9$ dikhai dega.) ✗ $[0, 2]$ par. Yeh step kyun? Hum haath se confirm karte hain jo IVT rule karne se mana kar diya — interval par maximum output $4$ hai, $9$ se neeche.

Verify: $max_{[0, 2]} x^{2} = 4$ aur $9 > 4$ , toh $x^{2} = 9$ ka $[0, 2]$ mein koi solution nahi; IVT theek se chup raha, aur yeh same-sign trap nahi hai.

Matrix ka recap

Recall Har cell ne kya sikhaya

Opposite signs (A, B) ::: IVT fire karta hai — ek root guaranteed hai. Specific $N$ (C) ::: pehle $g = f - N$ se shift karo, phir yeh Cell A ban jaata hai. Same sign (D) ::: IVT chup hai, "koi root nahi" NAHI — seedha check karo. Endpoint $N$ hit kare (E) ::: strict-between hi interior $c$ guarantee karta hai. Discontinuous (F) ::: hypothesis fail, koi guarantee nahi — jump check karo. Equal endpoints (G) ::: koi strictly-between $N$ nahi; IVT vacuous hai. Fixed point (H) ::: $x$ subtract karo, phir root dhoondo. Word problem (I) ::: motion model karo, sub-intervals mein split karo jo $N$ ko straddle karein. Existence vs count (J) ::: IVT kabhi count nahi karta — algebra ya Rolle use karo. $N$ band ke bahar (K) ::: IVT chup; value us interval par unreachable ho sakti hai.

Connections

Intermediate Value Theorem — woh parent rule jise yeh examples exercise karte hain.
Continuity — Cell F exactly isiliye fail hoti hai kyunki yeh toot jaati hai.
Bolzano's Theorem — Cells A, B, H, J ke peeche $N = 0$ engine.
Bisection Method — IVT ke promised $c$ ko actually locate karne ka tarika.
Rolle's Theorem aur Mean Value Theorem — counting/uniqueness ke tools (Cell J).
Fixed Point Theorems — Cell H ka natural ghar.