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. Consider this: 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. Now, 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:
The official docs gloss over this. That's a mistake That alone is useful..
[ (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 ).
People argue about this. Here's where I land on it.
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 ). Basically, you need to see to it 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
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = (g(x))^2 + 1 ]
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Substitute ( g(x) = 2x - 3 ):
[ f(g(x)) = (2x - 3)^2 + 1 ]
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Expand and simplify:
[ f(g(x)) = (4x^2 - 12x + 9) + 1 ]
[ f(g(x)) = 4x^2 - 12x + 10 ]
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Find ( g(f(x)) ):
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Replace ( x ) in ( g(x) ) with ( f(x) ):
[ g(f(x)) = 2(f(x)) - 3 ]
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Substitute ( f(x) = x^2 + 1 ):
[ g(f(x)) = 2(x^2 + 1) - 3 ]
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Expand and simplify:
[ g(f(x)) = 2x^2 + 2 - 3 ]
[ g(f(x)) = 2x^2 - 1 ]
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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 Simple as that..
Solution
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = \sqrt{g(x) - 2} ]
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Substitute ( g(x) = x + 5 ):
[ f(g(x)) = \sqrt{(x + 5) - 2} ]
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Simplify:
[ f(g(x)) = \sqrt{x + 3} ]
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Determine the domain of ( f(g(x)) ):
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The expression inside the square root must be non-negative:
[ x + 3 \geq 0 ]
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Solve for ( x ):
[ x \geq -3 ]
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Thus, the domain of ( f(g(x)) ) is ( x \geq -3 ), or in interval notation, ( [-3, \infty) ).
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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
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = \frac{1}{g(x)} ]
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Substitute ( g(x) = x - 1 ):
[ f(g(x)) = \frac{1}{x - 1} ]
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Find ( g(f(x)) ):
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Replace ( x ) in ( g(x) ) with ( f(x) ):
[ g(f(x)) = f(x) - 1 ]
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Substitute ( f(x) = \frac{1}{x} ):
[ g(f(x)) = \frac{1}{x} - 1 ]
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Simplify:
[ g(f(x)) = \frac{1 - x}{x} ]
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Determine the domain of ( f(g(x)) ):
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The denominator cannot be zero:
[ x - 1 \neq 0 ]
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Solve for ( x ):
[ x \neq 1 ]
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Thus, the domain of ( f(g(x)) ) is all real numbers except ( x = 1 ), or in interval notation, ( (-\infty, 1) \cup (1, \infty) ).
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Determine the domain of ( g(f(x)) ):
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The denominator cannot be zero:
[ x \neq 0 ]
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Thus, the domain of ( g(f(x)) ) is all real numbers except ( x = 0 ), or in interval notation, ( (-\infty, 0) \cup (0, \infty) ).
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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)) ) And that's really what it comes down to..
Solution
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = |g(x)| ]
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Substitute ( g(x) = x + 3 ):
[ f(g(x)) = |x + 3| ]
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Find ( g(f(x)) ):
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Replace ( x ) in ( g(x) ) with ( f(x) ):
[ g(f(x)) = f(x) + 3 ]
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Substitute ( f(x) = |x| ):
[ g(f(x)) = |x| + 3 ]
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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) ):
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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 ).
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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) ).
That's why, ( 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
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = \sin(g(x)) ]
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Substitute ( g(x) = 2x ):
[ f(g(x)) = \sin(2x) ]
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Find ( g(f(x)) ):
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Replace ( x ) in ( g(x) ) with ( f(x) ):
[ g(f(x)) = 2(f(x)) ]
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Substitute ( f(x) = \sin(x) ):
[ g(f(x)) = 2\sin(x) ]
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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
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Find ( f(g(h(x))) ):
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First, find ( g(h(x)) ):
[ g(h(x)) = h(x) + 1 = \sqrt{x} + 1 ]
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Now, find ( f(g(h(x))) ):
[ f(g(h(x))) = f(\sqrt{x} + 1) = (\sqrt{x} + 1)^2 ]
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Expand and simplify:
[ f(g(h(x))) = (\sqrt{x} + 1)(\sqrt{x} + 1) = x + 2\sqrt{x} + 1 ]
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Determine the domain of ( f(g(h(x))) ):
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Since ( h(x) = \sqrt{x} ), ( x ) must be non-negative:
[ x \geq 0 ]
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The domain of ( f(g(h(x))) ) is ( x \geq 0 ), or in interval notation, ( [0, \infty) ).
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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)) ) The details matter here..
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) ) That's the part that actually makes a difference..
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
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = e^{g(x)} ]
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Substitute ( g(x) = x^2 ):
[ f(g(x)) = e^{x^2} ]
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Find ( g(f(x)) ):
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Replace ( x ) in ( g(x) ) with ( f(x) ):
[ g(f(x)) = (f(x))^2 ]
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Substitute ( f(x) = e^x ):
[ g(f(x)) = (e^x)^2 ]
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Simplify:
[ g(f(x)) = e^{2x} ]
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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
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Find ( f(g(x)) ):
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Replace ( x ) in ( f(x) ) with ( g(x) ):
[ f(g(x)) = \frac{1}{\sqrt{g(x)}} ]
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Substitute ( g(x) = x - 2 ):
[ f(g(x)) = \frac{1}{\sqrt{x - 2}} ]
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Determine the domain of ( f(g(x)) ):
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The expression inside the square root must be positive (not zero, since it's in the denominator):
[ x - 2 > 0 ]
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Solve for ( x ):
[ x > 2 ]
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Thus, the domain of ( f(g(x)) ) is ( x > 2 ), or in interval notation, ( (2, \infty) ) Took long enough..
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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. Think about it: 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 confirm that the output of the inner function is a valid input for the outer function.
