Inverse functions — finding f⁻¹(x), horizontal line test

2.2.9 · Maths › Functions

Overview

Ek inverse function woh karta hai jo original function ne kiya tha usse undo karta hai. Agar $f$ , $x$ ko $y$ pe le jaata hai, toh $f^{- 1}$ , $y$ ko waapas $x$ pe le aata hai. Har function ka inverse nahi hota — sirf one-to-one (injective) functions ka hota hai. Horizontal line test yeh determine karta hai ki koi function one-to-one hai ya nahi, aur isliye invertible hai ya nahi.

Connections:

Function Composition — inverses ko $f (f^{- 1} (x)) = x$ ke zariye verify karna
Domain and Range — inverses ke liye domain/range swap hota hai
Bijective Functions — invertibility ke liye bijection zaroori hai
Exponential and Logarithmic Functions — classic inverse pair
Trigonometric Functions — invertibility ke liye domains restrict karna

Core Intuition

For example, agar $f (x) = 2 x + 3$ , $5 \to 13$ le jaata hai, toh $f^{- 1} (13) = 5$ . Inverse operation ko "rewind" karta hai.

LEKIN: har machine reversible nahi hoti. Agar do alag inputs ek hi output dete hain (jaise $f (x) = x^{2}$ jahan $f (2) = f (- 2) = 4$ ), toh tum usse uniquely reverse nahi kar sakte. Isliye hamen one-to-one condition ki zaroorat hai.

Figure — Inverse functions — finding f⁻¹(x), horizontal line test

Definitions and Core Concepts

Key property: $f^{- 1}$ ka graph, $f$ ka reflection hai line $y = x$ ke across.

Yeh kyun matter karta hai: Sirf one-to-one functions invertible hote hain. Agar $f (2) = f (5) = 7$ , toh $f^{- 1} (7)$ ambiguous hai (kya yeh 2 hai ya 5?). Koi ambiguity nahi matlab invertible.

Yeh kyun kaam karta hai: Ek horizontal line $y = c$ un saare points ko represent karti hai jo same $y$ -value rakhte hain. Agar woh graph ko do baar touch kare, toh do alag $x$ -values ek hi $y$ pe map ho rahe hain, jo one-to-one ki violation hai.

Inverse Derive Karna: The Swap Method

YEH process kyun kaam karti hai: Agar $(a, b)$ , $f$ ke graph pe hai (matlab $f (a) = b$ ), toh $(b, a)$ , $f^{- 1}$ ke graph pe hona chahiye (matlab $f^{- 1} (b) = a$ ). Equation mein $x$ aur $y$ swap karna literally input aur output ke roles swap karta hai.

First principles se derivation:

Hamen $f^{- 1}$ chahiye aisa ki $f (f^{- 1} (x)) = x$
Maano $f^{- 1} (x) = y$ . Toh $f (y) = x$ (inverse ki definition se)
Toh hamen $f (y) = x$ ko $y$ ke liye solve karna hai
Swap method bas ek notational trick hai is solving process ko organize karne ke liye

Worked Examples

Step 1: Likho $y = 3 x - 7$

Step 2: Swap karke milta hai $x = 3 y - 7$

Step 3: $y$ ke liye solve karo: $x + 7 = 3 y$ $y = \frac{x + 7}{3}$

Yeh step kyun? Hum $y$ ko isolate kar rahe hain taaki output ko input ke terms mein express kar sakein.

Step 4: Isliye $f^{- 1} (x) = \frac{x + 7}{3}$

Step 5: Verify karo: $f (f^{- 1} (x)) = f (\frac{x + 7}{3}) = 3 \cdot \frac{x + 7}{3} - 7 = (x + 7) - 7 = x ✓$ $f^{- 1} (f (x)) = f^{- 1} (3 x - 7) = \frac{( 3 x - 7 ) + 7}{3} = \frac{3 x}{3} = x ✓$

Verification kyun matter karta hai: Algebraic mistakes karna aasaan hai. Verification unhe pakad leta hai.

Step 1: Likho $y = \frac{2 x + 1}{x - 3}$

Step 2: Swap karke milta hai $x = \frac{2 y + 1}{y - 3}$

Step 3: $y$ ke liye solve karo: $x (y - 3) = 2 y + 1$ $x y - 3 x = 2 y + 1$ $x y - 2 y = 3 x + 1$ $y (x - 2) = 3 x + 1$ $y = \frac{3 x + 1}{x - 2}$

Yeh step kyun? Hum saare $y$ terms ek side collect karte hain taaki $y$ ko factor out kar sakein.

Step 4: Isliye $f^{- 1} (x) = \frac{3 x + 1}{x - 2}$ , $x \neq = 2$

Domain note: Denominator restriction badal jaati hai kyunki roles swap ho jaate hain. Original ki range limitation ( $y \neq = 2$ horizontal asymptote se) inverse ki domain restriction ban jaati hai.

Verification (abbreviated): $f (f^{- 1} (x)) = f (\frac{3 x + 1}{x - 2}) = \frac{2 \cdot \frac{3 x + 1}{x - 2} + 1}{\frac{3 x + 1}{x - 2} - 3} = \frac{\frac{6 x + 2 + x - 2}{x - 2}}{\frac{3 x + 1 - 3 x + 6}{x - 2}} = \frac{7 x}{7} = x ✓$

Horizontal Line Test: Draw karo $y = 4$ . Yeh parabola ko $x = - 2$ aur $x = 2$ pe intersect karta hai. Do intersections ka matlab hai not one-to-one.

Inverse kyun nahi: Agar hum $f^{- 1} (4)$ dhundhne ki koshish karte, toh do answers milte: $- 2$ aur $2$ . Ek inverse ko exactly ek output dena hota hai har input ke liye.

LEKIN: Agar hum domain restrict karte hain $x \geq 0$ pe, toh $f (x) = x^{2}$ , $[0, \infty)$ pe one-to-one ban jaata hai. Ab hum inverse dhundh sakte hain:

$y = x^{2}$ , $x \geq 0$
$x = y^{2}$ , $y \geq 0$
$y = x$ (sirf positive root)
$f^{- 1} (x) = x$

Restrict kyun karein? Domain restriction ambiguity hataa deti hai. Yeh technique trig functions ke liye crucial hai.

Step 1: $y = 3 x + 5 - 2$

Step 2: $x = 3 y + 5 - 2$

Step 3: $y$ ke liye solve karo: $x + 2 = 3 y + 5$ $(x + 2)^{3} = y + 5$ $y = (x + 2)^{3} - 5$

Dono sides ko cube kyun karein? Cubing, cube root ko undo karta hai, bilkul waise jaise inverse operation karna chahiye.

Step 4: $f^{- 1} (x) = (x + 2)^{3} - 5$

Domain/Range: Kyunki cube root saare real numbers ke liye defined hai aur one-to-one hai, dono $f$ aur $f^{- 1}$ ka domain aur range $R$ hai.

Common Mistakes & How to Fix Them

Yeh sahi kyun lagta hai: $- 1$ exponent notation reciprocals se liya gaya hai ( $x^{- 1} = 1/ x$ ).

Sach yeh hai: $f^{- 1}$ inverse function hai, reciprocal NAHI.

$f^{- 1} (f (x)) = x$ (composition identity deta hai)
$\frac{1}{f ( x )} \cdot f (x) = 1$ (multiplication 1 deta hai)

Example: $f (x) = 2 x$ ke liye:

Inverse: $f^{- 1} (x) = \frac{x}{2}$ , aur $f (f^{- 1} (x)) = 2 \cdot \frac{x}{2} = x$ ✓
Reciprocal: $\frac{1}{f ( x )} = \frac{1}{2 x}$ , jo bilkul alag hai

Fix: Hamesha apne inverse ko composition use karke verify karo, kabhi multiplication se nahi.

Yeh sahi kyun lagta hai: Algebra sahi hai.

Sach yeh hai: $f (x) = x - 2$ ka domain $x \geq 2$ aur range $y \geq 0$ hai. Jab hum swap karte hain, $f^{- 1}$ ka domain $x \geq 0$ hona chahiye (original range) aur range $y \geq 2$ (original domain).

Toh $f^{- 1} (x) = x^{2} + 2$ sirf $x \geq 0$ ke liye.

Fix: Domain/range swaps explicitly track karo:

Original domain → inverse range
Original range → inverse domain

Yeh sahi kyun lagta hai: Algorithm mechanical hai aur hamesha kuch na kuch produce karta hai.

Sach yeh hai: Agar $f$ one-to-one nahi hai, toh jo "inverse" tum dhundhhoge woh actually kaam nahi karega. $f (x) = x^{2}$ ke liye (saare reals):

Swap method deta hai $f^{- 1} (x) = \pm x$
Lekin yeh ek function nahi hai (ek input ke liye do outputs)!

Fix: Hamesha pehle horizontal line test lagao. Agar woh fail ho jaaye, toh inverse dhundhne se pehle domain restrict karo.

Horizontal Line Test — Geometric Proof

Horizontal line test kyun kaam karta hai?

Claim: $f$ one-to-one hai $⟺$ har horizontal line graph ko zyada se zyada ek baar intersect kare.

Proof ( $\Rightarrow$ ): Maano $f$ one-to-one hai. Contradiction ke liye maano ki ek horizontal line $y = c$ , graph ko do points $(x_{1}, c)$ aur $(x_{2}, c)$ pe intersect karti hai jahan $x_{1} \neq = x_{2}$ . Toh $f (x_{1}) = c$ aur $f (x_{2}) = c$ , toh $f (x_{1}) = f (x_{2})$ lekin $x_{1} \neq = x_{2}$ . Yeh one-to-one ka contradiction hai. ✓

Proof ( $\Leftarrow$ ): Maano har horizontal line zyada se zyada ek baar intersect karti hai. One-to-one dikhane ke liye, maano $f (x_{1}) = f (x_{2}) = c$ . Toh $(x_{1}, c)$ aur $(x_{2}, c)$ dono graph pe hain, toh woh horizontal line $y = c$ pe hain. Kyunki yeh line zyada se zyada ek baar intersect karti hai, hame $x_{1} = x_{2}$ milna chahiye. Isliye $f$ one-to-one hai. ✓

Domain and Range of Inverses

Derivation:

Maano $y \in Range (f)$ . Toh $y = f (x)$ kisi $x \in A$ ke liye. Inverse ki definition se, $f^{- 1} (y) = x$ , toh $y \in Domain (f^{- 1})$ .
Ulta, agar $y \in Domain (f^{- 1})$ , toh $f^{- 1} (y)$ defined hai, matlab koi $x$ exist karta hai aisa ki $f (x) = y$ , toh $y \in Range (f)$ .

Yeh kyun matter karta hai: Jab algebraically inverses dhundhte hain, toh domain restrictions ko sahi tarike se transfer karna zaroori hai.

Properties of Inverse Functions

Inverse of Inverse: $(f^{- 1})^{- 1} = f$ Derivation: Agar $g = f^{- 1}$ , toh $f (g (x)) = x$ aur $g (f (x)) = x$ . Iska matlab hai $f$ , $g$ ke inverse ki definition satisfy karta hai, toh $g^{- 1} = f$ .
Composition Property: $(f \circ g)^{- 1} = g^{- 1} \circ f^{- 1}$

Derivation: "Pehle $g$ karo phir $f$ karo" ko undo karne ke liye, hame "pehle $f$ undo karo phir $g$ undo karo" karna hoga (reverse order). Formally: $(g^{- 1} \circ f^{- 1}) (f \circ g) (x) = g^{- 1} (f^{- 1} (f (g (x))) = g^{- 1} (g (x)) = x$

Active Recall Questions

#flashcards/maths

What is the definition of an inverse function? :: Ek function $f^{- 1}$ aisa ki $f^{- 1} (f (x)) = x$ aur $f (f^{- 1} (x)) = x$ . Yeh woh karta hai jo $f$ ne kiya tha usse "undo" karta hai.

What condition must a function satisfy to have an inverse?

Function one-to-one (injective) hona chahiye: alag inputs alag outputs produce karein.

State the Horizontal Line Test :: Ek function one-to-one hai agar aur sirf agar har horizontal line uske graph ko zyada se zyada ek baar intersect kare.

What are the four steps to find an inverse algebraically?

1. Likho

y = f (x)

x

aur

y

swap karo 3.

y

ke liye solve karo 4.

y

f^{- 1} (x)

se replace karo

How are the domain and range of

f

and

f^{- 1}

related?

Domain of

f^{- 1}

= Range of

f

; Range of

f^{- 1}

= Domain of

f

. Woh swap ho jaate hain.

What is the difference between

f^{- 1} (x)

and

\frac{1}{f ( x )}

f^{- 1} (x)

inverse function hai (composition identity deta hai), jabki

\frac{1}{f ( x )}

reciprocal hai (multiplication 1 deta hai). Bilkul alag concepts hain.

How is the graph of

f^{- 1}

related to the graph of

f

f^{- 1}

ka graph,

f

ka reflection hai line

y = x

ke across.

Why can't

f (x) = x^{2}

have an inverse over all real numbers?

Kyunki yeh horizontal line test fail karta hai:

f (- 2) = f (2) = 4

, toh yeh one-to-one nahi hai. Multiple inputs ek hi output pe map ho rahe hain.

f (x) = 2 x - 5

, find

f^{- 1} (x)

y = 2 x - 5

, swap:

x = 2 y - 5

, solve:

y = \frac{x + 5}{2}

, toh

f^{- 1} (x) = \frac{x + 5}{2}

What is

(f \circ f^{- 1}) (x)

always equal to?

x

(identity function), inverse ki definition se.

Recall Feynman Technique: Explain to a 12-Year-Old

Socho tumhare paas ek secret code machine hai. Tum ek number daalo, aur woh ek coded number nikalta hai. For example, tumhari machine koi bhi number leta hai aur usse double karta hai, phir 3 add karta hai. Toh agar tum 5 daalo, tumhe 13 milta hai.

Ek inverse function ek decoder machine rakhne jaisa hai. Tum usse coded number (13) dete ho, aur woh tumhe original number (5) batata hai. Decoder opposite operations karta hai reverse order mein: pehle 3 subtract karo, phir 2 se divide karo.

Lekin yahan catch hai: har code machine ka decoder nahi hota! Agar tumhari machine kabhi kabhi alag numbers ke liye same code deti hai (jaise squaring: 3 aur -3 dono 9 dete hain), toh tum uniquely decode nahi kar sakte. Tumhe pata nahi hoga ki 9 aya 3 se ya -3 se.

Horizontal line test ek quick tarika hai check karne ka: apni machine ka graph draw karo, phir horizontal lines draw karo. Agar koi line graph ko do baar touch kare, tumhari machine decodable nahi hai (do alag inputs ne same output diya).

Decoder formula dhundhne ke liye: apni machine ka rule likhlo, input aur output swap karo, phir solve karo taaki input akela aa jaaye. Yahi tumhara decoder hai!

Horizontal line test ke liye: "Horizontal = One-to-One" — agar ek horizontal line ek baar hit kare, toh function one-to-one hai.

Summary

Inverse function $f^{- 1} (x)$ dhundhne ke liye:

Verify karo ki $f$ one-to-one hai (horizontal line test use karo)
Swap method use karo: $y = f (x) \to x = f (y) \to$ solve for $y$
Yaad rakho domain aur range swap hote hain
Hamesha composition use karke verify karo

Inverse function ko "undo" karta hai. Graphically, yeh $y = x$ ke across reflection hai. Sirf one-to-one functions invertible hote hain — horizontal line test is condition ko check karta hai yeh ensure karke ki har output exactly ek input se aata hai.