MathematicsRelations and FunctionsJEE

Relations and Functions · Part 8 of 20

Inverse Functions

An inverse runs a function backwards, and exists exactly for bijections. Find inverses algebraically, restrict domains, invert piecewise functions, use (gof)⁻¹ = f⁻¹og⁻¹, and solve f(x) = f⁻¹(x) on the line y = x.

Builds on: Part 7 · Bijections, Counting Functions and Composites.

Video coming soon

Graph

Mirror image in y = x

If (a, b) is on f, then (b, a) is on $f^{- 1}$ : the graphs are reflections in y = x. Range of f = domain of $f^{- 1}$ .

x² and √x reflected in y = x. — (2, 4) on x², (4, 2) on √x.

Parabola branches and their inverses. — Dashed curves are the inverses.

Restricting

Two branches of (x − 2)² + 1

On [2, ∞): $f^{- 1} (x) = 2 + x - 1$ ; on (−∞, 2]: $2 - x - 1$ . The sign must match the branch.

Piecewise

Invert piece by piece

f = x (x < 1), x² (1 ≤ x ≤ 4), 8√x (x > 4) has ranges (−∞, 1), [1, 16], (16, ∞). So $f^{- 1}$ = x, √x on [1, 16], x²/64 for x > 16.

Table of pieces and inverses. — New intervals are the old ranges.

Composition

$(g o f)^{- 1} = f^{- 1} o g^{- 1}$

Undo in reverse order. With f = 3x − 2 and $(g o f)^{- 1}$ = x − 2: $g^{- 1} = f \circ (g o f)^{- 1} = 3 x - 8$ , so g = (x + 8)/3.

Summary

Key formulas

f \circ f^{- 1} = I_{B}, f^{- 1} \circ f = I_{A}

Composition and inverses

(g \circ f)^{- 1} = f^{- 1} \circ g^{- 1}

Composition and inverses

Worked examples

From the video

1. f : R → [1, ∞), x² − 4x + 5. Largest intervals with an inverse, and the inverses.

(−∞, 2]: 2 − √(x − 1); [2, ∞): 2 + √(x − 1).

2. f(x) = (x − 2)/(x − 3). Find f⁻¹.

(3x − 2)/(x − 1).

3. f(x) = eˣ − e⁻ˣ. Find f⁻¹.

ln((x + √(x² + 4))/2).

4. Find a so that x² + 2ax + 1/16 = −a + √(a² + x − 1/16), x ≥ −a, has two distinct roots.

a ∈ [(2 − √5)/4, 1/4) ∪ (3/4, (2 + √5)/4]. (The book prints (√5 − 2)/4 at the left end, but a = 0 works.)

JEE-style question

Your turn

f : [2, ∞) → [1, ∞), f(x) = x² − 4x + 5. Then f⁻¹(x) is:

(a)2 + √(x − 1)

(b)2 − √(x − 1)

(c)2 ± √(x − 1)

(d)1/(x² − 4x + 5)

Show the answer and the traps

The branch is x ≥ 2, so x − 2 = +√(y − 1): f⁻¹(x) = 2 + √(x − 1). Option (a).

(b) is the inverse of the other branch, (−∞, 2]. (c) with ± is not even a function.

(d) is the reciprocal 1/f(x). f⁻¹ never means 1/f.

Watch out

Common mistakes

Choosing the wrong sign of the rootMatch the branch: x ≥ 2 needs + √.

Keeping the old domains when inverting piecesThe inverse lives on the ranges of the pieces.

Reading f⁻¹ as 1/ff⁻¹ undoes f; 1/f is the reciprocal.

Practice

Try these

1. Find the inverse of f(x) = (2x + 3)/(x − 1).

f⁻¹(x) = (x + 3)/(x − 2), x ≠ 2.

2. f : [0, ∞) → [−4, ∞), f(x) = x² − 4. Find f⁻¹.

√(x + 4).

3. f(x) = x³ + 1. Find f⁻¹(9).

4. f(x) = 2x + 3, g(x) = x − 1. Find (gof)⁻¹(x).

(x − 2)/2.