1.2 Composite Functions

1. Definition & Formation

A composite function joins two functions together. $f g (x)$ means we apply g first, then apply f to the result.

f g (x) = f (g (x))

Tip: Always read from right to left! "g acting on x, then f acting on that".

Visualizing the Process

Worked Examples : Definition & Formation

Given the functions

f (x) = 2 x - 3

and

g (x) = x^{2} + 1

, find the expressions for

f g (x)

and

g f (x)

1. Find fg(x): This means

f (g (x))

Replace

x

f

with

g (x)

f (g (x)) = 2 (g (x)) - 3

= 2 (x^{2} + 1) - 3

= 2 x^{2} + 2 - 3

= 2 x^{2} - 1

2. Find gf(x): This means

g (f (x))

Replace

x

g

with

f (x)

g (f (x)) = (f (x))^{2} + 1

= (2 x - 3)^{2} + 1

= (4 x^{2} - 12 x + 9) + 1

= 4 x^{2} - 12 x + 10

h (x) = \frac{1}{x + 2}

and

k (x) = \sqrt{x}

, find the value of

h k (4)

. Explain the order of operations.

Method: Work from the inside out. Evaluate

k (4)

first.

k (4) = \sqrt{4} = 2

Now find

h (2)

h (2) = \frac{1}{2 + 2} = \frac{1}{4} = 0.25

Order: 4 goes into k, output (2) goes into h.

Express the function

y = (3 x + 5)^{4}

as a composition of two simpler functions

f

and

g

where

y = f g (x)

We need an inner function

g

and outer function

f

Looking at the structure, the "inner" operation is

3 x + 5

Let

g (x) = 3 x + 5

The "outer" operation is raising to the power of 4.

Let

f (x) = x^{4}

Check:

f g (x) = f (3 x + 5) = (3 x + 5)^{4}

. Correct.

Given

f (x) = \sin (x)

and

g (x) = 2 x

, solve the equation

f g (x) = 0.5

for

0^{\circ} \leq x \leq 180^{\circ}

1. Form the composite:

f g (x) = f (2 x) = \sin (2 x)

2. Solve equation:

\sin (2 x) = 0.5

Let

u = 2 x

. Range for u:

0 \leq u \leq 360^{\circ}

\sin (u) = 0.5 ⟹ u = 30^{\circ}, 150^{\circ}

3. Find x:

2 x = 30^{\circ} ⟹ x = 15^{\circ}

2 x = 150^{\circ} ⟹ x = 75^{\circ}

Functions

f, g, h

are defined as

f (x) = x + 2

g (x) = 3 x

h (x) = x^{2}

. Find the expression for

h g f (x)

Work from right to left (inside out). First

f (x)

, then

g

, then

h

g f (x) = g (x + 2) = 3 (x + 2) = 3 x + 6

h (g f (x)) = h (3 x + 6)

= (3 x + 6)^{2}

Expanded:

9 x^{2} + 36 x + 36

Given

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

, find

f^{2} (x)

(which means

f f (x)

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

= \frac{(\frac{x + 1}{2}) + 1}{2}

Multiply top and bottom by 2 to clear fractions:

= \frac{x + 1 + 2}{4}

= \frac{x + 3}{4}

Functions

f

and

g

are defined by

f (x) = 2 x - 5

and

g (x) = x^{2} + 3

for

x \in R

(a) Find an expression for fg(x).

f g (x) = f (g (x)) = f (x^{2} + 3)

= 2 (x^{2} + 3) - 5

= 2 x^{2} + 6 - 5 = 2 x^{2} + 1

(b) Find an expression for gf(x).

g f (x) = g (f (x)) = g (2 x - 5)

= (2 x - 5)^{2} + 3

= (4 x^{2} - 20 x + 25) + 3 = 4 x^{2} - 20 x + 28

(c) Solve the equation gf(x) = 20.

4 x^{2} - 20 x + 28 = 20

4 x^{2} - 20 x + 8 = 0

Divide by 4:

x^{2} - 5 x + 2 = 0

Use formula:

x = \frac{5 \pm \sqrt{25 - 8}}{2} = \frac{5 \pm \sqrt{17}}{2}

The function

f

is defined by

f (x) = \frac{x}{x + 1}

x \neq - 1

, and

g (x) = 3 x - 1

(a) Find fg(x) in the form $\frac{a x + b}{c x + d}$ .

f g (x) = f (3 x - 1) = \frac{3 x - 1}{(3 x - 1) + 1}

= \frac{3 x - 1}{3 x}

(b) Solve the equation fg(x) = 1.

\frac{3 x - 1}{3 x} = 1

3 x - 1 = 3 x

- 1 = 0

This is a contradiction. There is no solution.

(c) State the value of x that must be excluded from the domain of fg.

Looking at

f g (x) = \frac{3 x - 1}{3 x}

, the denominator can't be zero.

x \neq 0

Given

f (x) = \cos x

and

g (x) = 2 x + \frac{π}{3}

for

0 \leq x \leq π

(a) Write down the expression for fg(x).

f g (x) = \cos (2 x + \frac{π}{3})

(b) Find the exact value of $f g (\frac{π}{6})$ .

f g (\frac{π}{6}) = \cos (2 (\frac{π}{6}) + \frac{π}{3})

= \cos (\frac{π}{3} + \frac{π}{3}) = \cos (\frac{2 π}{3})

= - 0.5

(c) Solve fg(x) = -0.5 for the given domain.

\cos (2 x + \frac{π}{3}) = - 0.5

Principal value (PV):

2 x + \frac{π}{3} = \frac{2 π}{3}, \frac{4 π}{3}

Case 1:

2 x + \frac{π}{3} = \frac{2 π}{3} ⟹ 2 x = \frac{π}{3} ⟹ x = \frac{π}{6}

Case 2:

2 x + \frac{π}{3} = \frac{4 π}{3} ⟹ 2 x = π ⟹ x = \frac{π}{2}

10.

Functions

f (x) = e^{2 x}

and

g (x) = \ln (x)

are given for

x > 0

(a) Find fg(x) in its simplest form.

f g (x) = e^{2 \ln (x)}

Use log laws:

2 \ln x = \ln (x^{2})

f g (x) = e^{\ln (x^{2})} = x^{2}

(b) Find gf(x) in its simplest form.

g f (x) = \ln (e^{2 x})

= 2 x \ln e = 2 x

(c) Are f and g inverse functions? Explain.

f and g are inverses only if

f g (x) = x

and

g f (x) = x

Here

f g (x) = x^{2}

and

g f (x) = 2 x

No, they are not inverses of each other (though related).

11.

Consider the function

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

for

x \neq 1

(a) Determine the expression for f^2(x) (i.e. ff(x)).

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

Multiply top and bottom by

(x - 1)

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

= \frac{x + 2 + 2 x - 2}{x + 2 - x + 1} = \frac{3 x}{3} = x

(b) Deduce the inverse function f^(-1)(x).

Since

f f (x) = x

, the function is its own inverse (self-inverse).

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

(c) Evaluate f^15(4).

Since

f^{2} (x) = x

, repeated application cycles:

f^{1} = f, f^{2} = I, f^{3} = f . . .

Odd powers are just f(x).

f^{15} (4) = f (4) = \frac{4 + 2}{4 - 1} = \frac{6}{3} = 2

12.

Given

f (x) = 3 x + k

and

g (x) = x^{2}

. The equation

f g (x) = g f (x)

has no real solutions for

x

. Find the range of possible values for k.

(a) Form the equation fg(x) = gf(x).

f g (x) = 3 (x^{2}) + k = 3 x^{2} + k

g f (x) = (3 x + k)^{2} = 9 x^{2} + 6 k x + k^{2}

Equation:

3 x^{2} + k = 9 x^{2} + 6 k x + k^{2}

(b) Rearrange into quadratic form.

0 = 6 x^{2} + 6 k x + (k^{2} - k)

(c) Use discriminant condition.

For "no real solutions", discriminant

D < 0

b^{2} - 4 a c = (6 k)^{2} - 4 (6) (k^{2} - k)

= 36 k^{2} - 24 k^{2} + 24 k

= 12 k^{2} + 24 k < 0

12 k (k + 2) < 0

Critical values at

k = 0, k = - 2

. Parabola opens up, so between roots.

Range:

- 2 < k < 0

2. Domain and Range

For $f g (x)$ to exist, the output of $g$ must be a valid input for $f$ .

Domain
The set of inputs $x$ into $g$ such that $g (x)$ is in the domain of $f$ .
Range
The set of final outputs from $f$ , generated by feeding the range of g into f.

Worked Examples : Domain and Range

Let

f (x) = \sqrt{x}

and

g (x) = x - 5

. Find the domain of the composite function

f g (x)

f g (x) = f (x - 5) = \sqrt{x - 5}

For the square root to be defined, the argument must be non-negative.

x - 5 \geq 0

x \geq 5

Note: The domain of

f g

is constrained by the requirement that the output of

g

must fit into the domain of

f

Given

f (x) = x^{2} - 2

with domain

x \in R

and

g (x) = \ln x

with domain

x > 0

. Find the range of

f g (x)

f g (x) = (\ln x)^{2} - 2

1. Consder the range of inner function

g (x) = \ln x

. Since domain is

x > 0

\ln x

can take any real value

(- \infty, \infty)

2. Let

u = \ln x

. We need range of

u^{2} - 2

where

u \in R

3. Minimum value of

u^{2}

is 0.

4. Minimum value of

u^{2} - 2

- 2

Range:

f g (x) \geq - 2

Let

f (x) = \frac{1}{x}

and

g (x) = x - 2

. Explain why

f g (2)

is undefined.

Check the inner function first:

g (2) = 2 - 2 = 0

Now try to pass this to

f

f (0) = \frac{1}{0}

Division by zero is undefined.

Therefore,

x = 2

must be excluded from the domain of

f g (x)

Given

f (x) = 2 x + 1, x \geq 0

and

g (x) = x^{2}, x \in R

. Find the range of

g f (x)

g f (x) = g (2 x + 1) = (2 x + 1)^{2}

1. Range of inner function

f

: Since

x \geq 0

2 x + 1 \geq 1

2. Now consider

g (u) = u^{2}

for domain

u \geq 1

3. Since

u^{2}

is increasing for positive

u

, the minimum is at

^{2} = 1

Range:

g f (x) \geq 1

Determine the domain of

f (x) = \ln (1 - x^{2})

by viewing it as a composition.

Inner function

g (x) = 1 - x^{2}

. Outer

h (u) = \ln u

h (u)

requires

u > 0

So we need

1 - x^{2} > 0

x^{2} < 1

- 1 < x < 1

Domain:

(- 1, 1)

f (x) = e^{x}

and

g (x) = \sqrt{x}

, find the range of

f g (x)

if the domain of

g

x \geq 0

f g (x) = e^{\sqrt{x}}

Since

x \geq 0

\sqrt{x} \geq 0

The exponential function

e^{u}

is strictly increasing.

Minimum value is when

\sqrt{x} = 0

, giving

e^{0} = 1

x \to \infty

e^{\sqrt{x}} \to \infty

Range:

f g (x) \geq 1

(or

[1, \infty)

3. Non-Commutativity

Order matters! In general, $f g (x) \neq g f (x)$ .

Think of it like putting on socks and shoes.
Socks then Shoes: Result is dressed feet.
Shoes then Socks: Result is socks over shoes (very different!).

exceptions exist, e.g., if

f (x) = x

or if functions are inverses of each other.

Worked Examples : Non-Commutativity

Let

f (x) = 3 x

and

g (x) = x + 2

. Show algebraically that

f g (x) \neq g f (x)

f g (x) = f (x + 2) = 3 (x + 2) = 3 x + 6

g f (x) = g (3 x) = 3 x + 2

Comparing the terms:

3 x + 6 \neq 3 x + 2

(unless

0 = 4

, impossible).

Thus composition is not commutative in this case.

Find the value of

k

for which

f g (x) = g f (x)

for all

x

, given

f (x) = 2 x + 1

and

g (x) = 3 x + k

1. Find

f g (x)

2 (3 x + k) + 1 = 6 x + 2 k + 1

2. Find

g f (x)

3 (2 x + 1) + k = 6 x + 3 + k

3. Equate coefficient-free terms (constants):

2 k + 1 = 3 + k

4. Solve for k:

k = 2

Given

f (x) = x^{2}

and

g (x) = x - 1

, solve the equation

f g (x) = g f (x)

f g (x) = (x - 1)^{2} = x^{2} - 2 x + 1

g f (x) = (x^{2}) - 1 = x^{2} - 1

Set equal:

x^{2} - 2 x + 1 = x^{2} - 1

Subtract

x^{2}

from both sides:

- 2 x + 1 = - 1

- 2 x = - 2

x = 1

Explain with an example why

f g (x)

might exist even if

g f (x)

does not.

Let

f (x) = \sqrt{x}

(Domain

x \geq 0

) and

g (x) = x^{2}

(Domain

R

f g (x) = \sqrt{x^{2}} = | x |

. Defined for all

x

Consider a different case:

f (x) = x + 5, g (x) = - x

. This is bad example.

Better example:

f (x) = \ln x

and

g (x) = - x^{2} - 1

f (g (x)) = \ln (- x^{2} - 1)

. Since

- x^{2} - 1

is always negative, this is never defined.

However,

g (f (x)) = - (\ln x)^{2} - 1

exists for

x > 0

Under what specific condition is

f g (x) = g f (x) = x

for all

x

This strictly happens when

g

is the inverse function of

f

, denoted as

f^{- 1}

g (x) = f^{- 1} (x)

, then applying them in sequence returns the original input.

Given

f (x) = m x + c

and

g (x) = n x + d

. Find the condition on m, n, c, d for commutativity (

f g = g f

f g (x) = m (n x + d) + c = m n x + m d + c

g f (x) = n (m x + c) + d = n m x + n c + d

Slopes

m n

match. We need intercepts to match:

m d + c = n c + d

m d - d = n c - c

d (m - 1) = c (n - 1)

This is the condition. Useful special cases: if

m = 1, n = 1

, it always holds.