Composition of functions

COMPOSITION OF FUNCTIONS

In general, given any two functions f and g, we start with a number x in the domain of g and find its image g(x). If this number g(x) is in the domain of f, then we can calculate the value of f (g(x)). The result is a new function h(x) = f(g(x)) obtained by substituting g into f. It is called the composition (or composite) of f and g and is denoted by fog ("f circle g").

Definition: Given two functions f and g, the composite function f og (also called the composition of ƒ and g) is defined by

(fog)(x) = f(g(x))

The domain of f og is the set of all x in the domain of g such that g(x) is in the domain of f. In other words, (f og) (x) is defined whenever both g(x) and f(g(x)) are defined. Figure 11 shows how to picture f o g in terms of machines.

It is possible to take the composition of three or more functions. For instance, the composite function f ogoh is found by first applying h, then g, and then ƒ as follows:

 (f°g o h)(x) = ƒ (g(h(x)))

PROBLEMS UNDER COMPOSITION OF FUNCTIONS

Example 17. If f(x) = x2 and g(x) = x − 3 find the composite functions fog and gof.

Solution:

We have

(f og)(x) = f(g(x)) = f(x − 3) = (x − 3)2

 (g o f)(x) = g(f(x)) = g(x2) = x2 − 3

NOTE: You can see from above Example that, in general, fog ≠ gof. Remember, the notation fog means that the function g is applied first and then ƒ is applied second. In the above Example, fog is the function that first subtracts 3 and then squares; gof is the function that first squares and then subtracts 3.

Example 18. Let f(x) = x2 + 3 and g(x) = √x. Find

(a) (f og)(x)

(b) (g o f)(x)

Solution: (a). The formula for f(g(x)) is

ƒ(g(x)) = [g(x)]2 + 3 = (√x)2 + 3 = x + 3.

Since the domain of g is [0, +∞) and the domain of ƒ is (‒∞, ∞), the domain of fog consists of all x in [0, +∞) such that g(x) = √x lies in (‒∞, +∞); thus, the domain of fog is [0, +∞). Therefore,

(f og) (x) = x + 3, x≥0

(b). The formula for g(f(x)) is

g(f(x)) = √f(x) = √[x2 + 3]

Since the domain of f is (‒∞, +∞) and the domain of g is [0, +∞), the domain of gof consists of all x in (‒∞, +∞) such that f(x) = x2+ 3 lies in [0, +∞). Thus, the domain of gof is (‒∞, +∞). Therefore,

(g o f)(x) = √[x2+3]

There is no need to indicate that the domain is (‒∞, +∞), since this is the natural domain of √[x2 + 3].

Example 19. Express sin(x3) a composition of two functions

Solution: To evaluate sin(x3) we would first compute x3 and then take the sine, so g(x) = x3 is the inside function and f(x) = sin x the outside function.

Therefore, sin(x3) = f(g(x))

Example 20. If f(x) = √x and g(x) = √[2‒x], find each function and its domain.

(a) fog (b) gof (c) fof (d) gog

Solution:

(a) (fog) (x) = f(g(x)) = f(√[2 –x])) = √√[2‒x] = 4√[2‒x]

The domain of fog is { x|2‒x ≥ 0} = {x|x ≤ 2} = (‒∞,2].

(b) (g ° f) (x) = g(f(x)) = g(√x)) = √√[2 ‒ √x]

For √x to be defined we must have x ≥0. For √[2‒√x] to be defined we must have 2‒√x ≥ 0, that is, √x ≤ 2 or x ≤ 4. Thus 0 ≤ x ≤ 4, so the domain of g o f is the closed interval [0,4].

(c) (f°f) (x) = f(f(x)) = ƒ(√x) = √√x = 4√x

The domain of f o f is [0, ∞).

(d) (g o g) (x) = g(g(x)) = g(√[2 –x]) = √[2 ‒ √[2‒x] ]

This expression is defined when both 2‒x ≥ 0 and 2 ‒ √[2‒x] ≥ 0. The first inequality means x ≤ 2, and the second is equivalent to √[2‒x] ≤ 2, or 2‒x ≤ 4 or x ≥ − 2. Thus ‒2 ≤ x ≤ 2 so the domain of gog is the closed interval [−2,2].

Example 21. Find (fogoh)(x) if

f(x)=√x, g(x) = 1/x, h(x) = x3

Solution:

(f ogoh)(x) = f(g(h(x))) = f(g(x3))

Example 22. Find f o g o h if f(x) = x / (x + 1). g(x) = x10. and h(x)= x+3.

Solution: (fogoh)(x) = f(g(h(x))) = f(g(x+3))

= f((x+3)10) = [ (x+3)10 ] / [ (x+3)10 + 1 ]

So far we have used composition to build complicated functions from simpler ones. But in calculus it is often useful to be able to decompose a complicated function into simpler ones, as in the following example.

Example 23. Given F(x) = cos2(x+9) find functions f, g, and h such that F = fogoh.

Solution: Since F(x)= [cos(x+9)]2 the formula for F says: First add 9, then take the cosine of the result, and finally square. So we let

h(x) = x+9

g(x) = cos x

f(x) = x2

Then, (f o g o h)(x) = f (g(h(x))) = f(g(x + 9)) = f(cos(x + 9))

 [cos(x+9)]2 = F(x)

Example 24. Use the table to evaluate each expression.

(a) f(g(1)) (b) g(f(1)) (c) f(f(1)) (d)g(g(1)) (e) (gof)(3) (f) (fog)(6)

Solution:

a) f(g(1)) = ƒ(6) = 5

b) g(f(1)) = g(3) = 2

c) ƒ(ƒ(1)) = ƒ(3) = 4

d) g(g(1)) = g(6) = 3

e) (gof)(3) = g(f(3)) = g(4) = 1

f) (fog)(6) = f(g(6)) = f(3) = 4

EXERCISE

3. Find f+g,f‒g.fg and f/g and state their domains.

(a) f(x) = x3 + 2x2, g(x) = 3x2‒1

(b) f(x)=√[3‒x], g(x) = √[x2‒1]

4. Find the functions (a) fog. (b) gof (c) f of and (d) g o g and their domains.

(a) f(x) = x2 ‒ 1, g(x) = 2x + 1

(b) f(x) = x ‒2, g(x) = x2 + 3x + 4

(c) f(x) = 1‒3x, g(x) = cos x

(d) f(x) = √x, g(x) = 3√[1‒x]

(e) f(x) = x + 1/x, g(x) = x+1 / x+2

(f) f(x) = x / 1+x , g(x) = sin 2x

5. Find fogoh.

(a) f(x) = x+1, g(x) = 2x, h(x) = x ‒ 1

(b) f(x) = 2x − 1, g(x) = x2, h(x) = 1 ‒ x

(c) f(x) = √[x‒3], g(x) = x2, h(x) = x2 + 2

(d) f(x) = tanx, g(x) = x / x‒1, h(x) = 3√x

6. Express the function in the form fog.

(a) F(x) = (x2 + 1)10

(b) F(x) = sin(√x)

(c) F(x) = 3√x / 1+3√x

(d) G(x) = 3√ [ x / 1+x ]

(e) u(t) = √cost

(f) u(t) = tant / 1+tant

7. Express the function in the form fogoh.

(a) H(x) = 1‒3x2

(b) H(x) = 8√[ 2 + |x| ]

(c) H(x) = sec4(√x)