MathematicsRelations and FunctionsJEE

Relations and Functions · Part 5 of 20

Algebra of Functions and Composite Functions

Combine functions with +, −, ×, ÷ and composition, always asking what the new domain is. The heart of this part is the composite of two piecewise functions, done by graphing the inner function and reading off bands.

Builds on: Part 4 · Functions: Definition, Intervals and Domain.

Video coming soon

Algebra of functions

Both must exist

f ± g and fg live on $D_{f} \cap D_{g}$ ; f/g also drops the zeros of g. With f = √x and g = √(1 − x): f + g on [0, 1], f/g on [0, 1).

Number lines for the two domains. — Intersection first, then remove g = 0.

Composite function definition. — Output of f must land in the domain of g.

Composition

$g o f (x) = g (f (x))$

$D_{g o f} = {x \in D_{f} : f (x) \in D_{g}}$ . With f = x − 2 and g = √x: gof = √(x − 2) on [2, ∞) but fog = √x − 2 on [0, ∞). Order matters.

Piecewise composite

Graph g, read off the bands

f(x) = x + 1 (x ≤ 1), 2x + 1 (1 < x ≤ 2); g(x) = x² (−1 ≤ x < 2), x + 2 (2 ≤ x ≤ 3). fog uses the band g ≤ 1 for x ∈ [−1, 1] and 1 < g ≤ 2 for x ∈ (1, √2]; everything else lands outside $D_{f}$ .

Graph of g with two shaded bands. — Blue band: g ≤ 1 · orange band: 1 < g ≤ 2.

Graph of the composite with its range. — Range marked on the y-axis.

Result

fog: domain [−1, √2]

fog = x² + 1 on [−1, 1] and 2x² + 1 on (1, √2]. Range [1, 2] ∪ (3, 5]. The book prints [0, 2] ∪ (3, 5], but x² + 1 is never below 1.

Worked examples

From the video

1. fog for the piecewise f and g above: formula, domain, range.

x² + 1 on [−1, 1], 2x² + 1 on (1, √2]; domain [−1, √2]; range [1, 2] ∪ (3, 5].

2. g(x) = x² + x − 1, gof(x) = 4x² − 10x + 5. Find f(5/4).

gof(5/4) = −5/4, so t² + t + 1/4 = 0 and f(5/4) = −1/2.

JEE-style question

Your turn

f(x) = √x and g(x) = x² − 1. The domain of fog is:

(a)R

(b)[1, ∞)

(c)(−∞, −1] ∪ [1, ∞)

(d)[−1, 1]

Show the answer and the traps

fog(x) = √(x² − 1) needs x² ≥ 1: x ≤ −1 or x ≥ 1. Option (c).

R is the domain of g, not of fog. [1, ∞) forgets the negative side.

[−1, 1] solves x² − 1 ≤ 0: the inequality the wrong way round.

Watch out

Common mistakes

Taking the domain of fog as the domain of gfog also needs g(x) inside D_f; here [−1, 3] shrinks to [−1, √2].

Keeping cases where the two conditions share no xg = x + 2 ≤ 1 needs x ≤ −1 and x ≥ 2: drop it.

Assuming fog = gof√(x − 2) and √x − 2 differ in formula and domain.

Practice

Try these

1. f(x) = √(x − 1), g(x) = √(5 − x). Domains of f + g and f/g?

[1, 5] and [1, 5).

2. f(x) = 2x + 1, g(x) = x². Find fog and gof.

fog = 2x² + 1; gof = (2x + 1)².

3. f(x) = 1/(x − 1). Find fof and its domain.

(x − 1)/(2 − x); domain R − {1, 2}.

4. f(x) = √x, g(x) = 1 − x². Domain of fog?

[−1, 1].