2.7 Composition Of Functions Practice Set 1

The composition of functions, a fundamental concept in mathematics, involves combining two functions in a way that the output of one function becomes the input of another. This operation, denoted as ( f(g(x)) ), allows us to create more complex functions and is widely used in various fields, including calculus, computer science, and engineering. Mastering the composition of functions requires a solid understanding of function notation, domain, and range, and the ability to perform algebraic manipulations. In this comprehensive practice set, we will explore various aspects of function composition through detailed examples and explanations.

Understanding Function Composition

Before diving into the practice set, let’s clarify the basic concepts of function composition.

Definition

Given two functions ( f(x) ) and ( g(x) ), the composition of ( f ) with ( g ), denoted as ( f(g(x)) ), is defined as applying the function ( g ) to ( x ) first, and then applying the function ( f ) to the result. In other words:

[ (f \circ g)(x) = f(g(x)) ]

Notation

The notation ( f(g(x)) ) is read as "f of g of x." It means that the function ( g ) is applied to ( x ), and the result is then used as the input for the function ( f ).

Domain and Range

The domain of the composite function ( f(g(x)) ) is the set of all ( x ) in the domain of ( g ) such that ( g(x) ) is in the domain of ( f ). In other words, you need to ensure that the output of ( g(x) ) can be a valid input for ( f(x) ).

The range of ( f(g(x)) ) is the set of all possible outputs of ( f ) when the inputs are the outputs of ( g ).

Key Points

• Order Matters: In general, ( f(g(x)) ) is not the same as ( g(f(x)) ). Function composition is not commutative.
• Inner and Outer Functions: In ( f(g(x)) ), ( g(x) ) is the inner function, and ( f(x) ) is the outer function.
• Domain Restrictions: Pay close attention to domain restrictions for both the inner and outer functions.

Practice Set 1: Composition of Functions

Let's work through a variety of examples to solidify your understanding of function composition.

Example 1: Basic Composition

Given:

[ f(x) = x^2 + 1 ]

[ g(x) = 2x - 3 ]

Find ( f(g(x)) ) and ( g(f(x)) ).

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = (g(x))^2 + 1 ]

   • Substitute ( g(x) = 2x - 3 ):

     [ f(g(x)) = (2x - 3)^2 + 1 ]

   • Expand and simplify:

     [ f(g(x)) = (4x^2 - 12x + 9) + 1 ]

     [ f(g(x)) = 4x^2 - 12x + 10 ]

2. Find ( g(f(x)) ):

   • Replace ( x ) in ( g(x) ) with ( f(x) ):

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

   • Substitute ( f(x) = x^2 + 1 ):

     [ g(f(x)) = 2(x^2 + 1) - 3 ]

   • Expand and simplify:

     [ g(f(x)) = 2x^2 + 2 - 3 ]

     [ g(f(x)) = 2x^2 - 1 ]

Conclusion:

• ( f(g(x)) = 4x^2 - 12x + 10 )
• ( g(f(x)) = 2x^2 - 1 )

Example 2: Composition with Radicals

Given:

[ f(x) = \sqrt{x - 2} ]

[ g(x) = x + 5 ]

Find ( f(g(x)) ) and determine its domain.

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = \sqrt{g(x) - 2} ]

   • Substitute ( g(x) = x + 5 ):

     [ f(g(x)) = \sqrt{(x + 5) - 2} ]

   • Simplify:

     [ f(g(x)) = \sqrt{x + 3} ]

2. Determine the domain of ( f(g(x)) ):

   • The expression inside the square root must be non-negative:

     [ x + 3 \geq 0 ]

   • Solve for ( x ):

     [ x \geq -3 ]

   • Thus, the domain of ( f(g(x)) ) is ( x \geq -3 ), or in interval notation, ( [-3, \infty) ).

Conclusion:

• ( f(g(x)) = \sqrt{x + 3} )
• The domain of ( f(g(x)) ) is ( [-3, \infty) ).

Example 3: Composition with Rational Functions

Given:

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

[ g(x) = x - 1 ]

Find ( f(g(x)) ) and ( g(f(x)) ), and determine their domains.

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = \frac{1}{g(x)} ]

   • Substitute ( g(x) = x - 1 ):

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

2. Find ( g(f(x)) ):

   • Replace ( x ) in ( g(x) ) with ( f(x) ):

     [ g(f(x)) = f(x) - 1 ]

   • Substitute ( f(x) = \frac{1}{x} ):

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

   • Simplify:

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

3. Determine the domain of ( f(g(x)) ):

   • The denominator cannot be zero:

     [ x - 1 \neq 0 ]

   • Solve for ( x ):

     [ x \neq 1 ]

   • Thus, the domain of ( f(g(x)) ) is all real numbers except ( x = 1 ), or in interval notation, ( (-\infty, 1) \cup (1, \infty) ).

4. Determine the domain of ( g(f(x)) ):

   • The denominator cannot be zero:

     [ x \neq 0 ]

   • Thus, the domain of ( g(f(x)) ) is all real numbers except ( x = 0 ), or in interval notation, ( (-\infty, 0) \cup (0, \infty) ).

Conclusion:

• ( f(g(x)) = \frac{1}{x - 1} ) with domain ( (-\infty, 1) \cup (1, \infty) )
• ( g(f(x)) = \frac{1 - x}{x} ) with domain ( (-\infty, 0) \cup (0, \infty) )

Example 4: Composition with Absolute Value

Given:

[ f(x) = |x| ]

[ g(x) = x + 3 ]

Find ( f(g(x)) ) and ( g(f(x)) ).

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = |g(x)| ]

   • Substitute ( g(x) = x + 3 ):

     [ f(g(x)) = |x + 3| ]

2. Find ( g(f(x)) ):

   • Replace ( x ) in ( g(x) ) with ( f(x) ):

     [ g(f(x)) = f(x) + 3 ]

   • Substitute ( f(x) = |x| ):

     [ g(f(x)) = |x| + 3 ]

Conclusion:

• ( f(g(x)) = |x + 3| )
• ( g(f(x)) = |x| + 3 )

Example 5: Composition with Piecewise Functions

Given:

[ f(x) = \begin{cases} x^2, & \text{if } x \geq 0 \ 2x, & \text{if } x < 0 \end{cases} ]

[ g(x) = x - 1 ]

Find ( f(g(x)) ).

Solution

To find ( f(g(x)) ), we need to consider the two cases of ( f(x) ):

1. Case 1: ( g(x) \geq 0 )

   • If ( g(x) \geq 0 ), then ( x - 1 \geq 0 ), which means ( x \geq 1 ).
   • In this case, ( f(g(x)) = (g(x))^2 = (x - 1)^2 ).

2. Case 2: ( g(x) < 0 )

   • If ( g(x) < 0 ), then ( x - 1 < 0 ), which means ( x < 1 ).
   • In this case, ( f(g(x)) = 2(g(x)) = 2(x - 1) ).

Therefore, ( f(g(x)) ) is defined as:

[ f(g(x)) = \begin{cases} (x - 1)^2, & \text{if } x \geq 1 \ 2(x - 1), & \text{if } x < 1 \end{cases} ]

Conclusion:

[ f(g(x)) = \begin{cases} (x - 1)^2, & \text{if } x \geq 1 \ 2(x - 1), & \text{if } x < 1 \end{cases} ]

Example 6: Composition with Trigonometric Functions

Given:

[ f(x) = \sin(x) ]

[ g(x) = 2x ]

Find ( f(g(x)) ) and ( g(f(x)) ).

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = \sin(g(x)) ]

   • Substitute ( g(x) = 2x ):

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

2. Find ( g(f(x)) ):

   • Replace ( x ) in ( g(x) ) with ( f(x) ):

     [ g(f(x)) = 2(f(x)) ]

   • Substitute ( f(x) = \sin(x) ):

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

Conclusion:

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

Example 7: Advanced Composition with Multiple Functions

Given:

[ f(x) = x^2 ]

[ g(x) = x + 1 ]

[ h(x) = \sqrt{x} ]

Find ( f(g(h(x))) ) and determine its domain.

Solution

1. Find ( f(g(h(x))) ):

   • First, find ( g(h(x)) ):

     [ g(h(x)) = h(x) + 1 = \sqrt{x} + 1 ]

   • Now, find ( f(g(h(x))) ):

     [ f(g(h(x))) = f(\sqrt{x} + 1) = (\sqrt{x} + 1)^2 ]

   • Expand and simplify:

     [ f(g(h(x))) = (\sqrt{x} + 1)(\sqrt{x} + 1) = x + 2\sqrt{x} + 1 ]

2. Determine the domain of ( f(g(h(x))) ):

   • Since ( h(x) = \sqrt{x} ), ( x ) must be non-negative:

     [ x \geq 0 ]

   • The domain of ( f(g(h(x))) ) is ( x \geq 0 ), or in interval notation, ( [0, \infty) ).

Conclusion:

• ( f(g(h(x))) = x + 2\sqrt{x} + 1 )
• The domain of ( f(g(h(x))) ) is ( [0, \infty) ).

Example 8: Decomposing Functions

Decompose the function ( H(x) = \sqrt{x^2 + 1} ) into two functions ( f(x) ) and ( g(x) ) such that ( H(x) = f(g(x)) ).

Solution

One possible decomposition is:

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

Then, ( f(g(x)) = f(x^2 + 1) = \sqrt{x^2 + 1} = H(x) ).

Another possible decomposition is:

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

Then, ( f(g(x)) = f(x^2) = \sqrt{x^2 + 1} = H(x) ).

Conclusion:

• One possible decomposition is ( f(x) = \sqrt{x} ) and ( g(x) = x^2 + 1 ).
• Another possible decomposition is ( f(x) = \sqrt{x + 1} ) and ( g(x) = x^2 ).

Example 9: Composition with Exponential Functions

Given:

[ f(x) = e^x ]

[ g(x) = x^2 ]

Find ( f(g(x)) ) and ( g(f(x)) ).

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = e^{g(x)} ]

   • Substitute ( g(x) = x^2 ):

     [ f(g(x)) = e^{x^2} ]

2. Find ( g(f(x)) ):

   • Replace ( x ) in ( g(x) ) with ( f(x) ):

     [ g(f(x)) = (f(x))^2 ]

   • Substitute ( f(x) = e^x ):

     [ g(f(x)) = (e^x)^2 ]

   • Simplify:

     [ g(f(x)) = e^{2x} ]

Conclusion:

• ( f(g(x)) = e^{x^2} )
• ( g(f(x)) = e^{2x} )

Example 10: Finding the Domain with Multiple Restrictions

Given:

[ f(x) = \frac{1}{\sqrt{x}} ]

[ g(x) = x - 2 ]

Find ( f(g(x)) ) and determine its domain.

Solution

1. Find ( f(g(x)) ):

   • Replace ( x ) in ( f(x) ) with ( g(x) ):

     [ f(g(x)) = \frac{1}{\sqrt{g(x)}} ]

   • Substitute ( g(x) = x - 2 ):

     [ f(g(x)) = \frac{1}{\sqrt{x - 2}} ]

2. Determine the domain of ( f(g(x)) ):

   • The expression inside the square root must be positive (not zero, since it's in the denominator):

     [ x - 2 > 0 ]

   • Solve for ( x ):

     [ x > 2 ]

   • Thus, the domain of ( f(g(x)) ) is ( x > 2 ), or in interval notation, ( (2, \infty) ).

Conclusion:

• ( f(g(x)) = \frac{1}{\sqrt{x - 2}} )
• The domain of ( f(g(x)) ) is ( (2, \infty) ).

Conclusion

The composition of functions is a powerful tool in mathematics that allows us to create complex functions from simpler ones. By understanding the basic definitions, notation, and domain restrictions, you can effectively work with composite functions in various contexts. This practice set has provided a range of examples, from basic compositions to more complex scenarios involving radicals, rational functions, absolute values, piecewise functions, trigonometric functions, and exponential functions. By working through these examples and understanding the solutions, you can build a strong foundation in function composition. Remember to always pay attention to the domain of the inner function and ensure that the output of the inner function is a valid input for the outer function.