The concept of a function is essential in mathematics. There are two common notations in use:
(a) | ƒ(x) = x^{2} + 2,_{ } | |
and | (b) | ƒ : x x^{2} + 2._{ } |
Form (a) is commonly used. Form (b) is interpreted as:
the function ƒ maps x to x^{2} + 2 |
Example 1: If two functions are given as ƒ(x) = 2x + 3 and g(x) = 3 − x^{2} , then:
Example 2: Find the numbers which map to zero under the function:
Solution: The function can also be written as h(x) = x^{2} − 9 and if x maps to zero then h(x) = 0, i.e.:
x^{2} − 9_{ } | = | 0 |
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x^{2}_{ } | = | 9 |
The numbers are 3 and −3, as when squared they both equal 9.
Exercise 1: Two functions are given as: h : x x^{2} − 4 and g : x 10x + 5 . Find the following:
Solution: The function is h(x) = x^{2} − 4 so:
h(1) | = | 1^{2} − 4_{ } |
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= | 1 − 4 = −3 |
Solution: The function is h(x) = x^{2} − 4 so:
h(-2) | = | -2^{2} − 4_{ } |
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= | 4 − 4 = 0 |
Solution:The function is h(x) = x^{2} − 4 so:
h(0) | = | (3)^{2} − 4_{ } |
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= | 0 − 4 = −4 |
Solution: The function is g(x) = 10x + 5 so:
g(3) | = | 10 × (3) + 5 |
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= | 30 + 5 = 35 |
Solution: The function is g(x) = 10x + 5 so:
g(−1) | = | 10 × (−1) + 5 |
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= | −10 + 5 = −5 |
Solution: If h(x) = 12 , then since h(x) = x^{2} − 4 :
x^{2} − 4_{ } | = | 12 |
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x^{2}_{ } | = | 12 + 4 = 16 |
x | = | ±4 |
Click on questions to reveal their solutions
The function ƒ(x) = 2x + 3 in Example 1 may be represented as a flow diagram:
From the diagram it is clear that the order of the operations cannot be confused: first multiply by 2 and then add 3.
Example 3: Draw a ﬂow diagram for each of the following functions:
Solution:
Solution:
The exercise below is designed to give you some practice at addition: and subtraction of fractions.
Exercise 2: Draw flow diagrams for each of the following functions:
Solution: For h : x 6x + 1 the flow diagram is:
Solution: For h : x 4(3 − 2x) the flow diagram is:
Solution: For h : x (2x − 5)^{2} the flow diagram is:
Solution:
Solution:
Solution:
Click on questions to reveal their solutions
Now try this short quiz:
Quiz 1: From the functions listed below choose the one which is given by the following ﬂow diagram:
Explanation: The flow diagram:
is completed as:
- which is the function:
k : ( x^{3} 4 − 1 ) ^{2} + 4The function ƒ(x) = 2x + 3 , from Example 1, is composed of two simpler functions, i.e. multiply by 2 and add 3. If these two functions are written as h : 2x and g : x + 3 then the composition of these two functions is written gh (sometimes as gοh or g (h (x))) .
Example 4: If h : 2x^{2} and g : √x + 5 , find the composite function gh
Solution: Applying first h and then g results in the composite function:
This can best be seen by using a flow diagram:
x | square | x^{2} | multiply by 2 | 2x^{2} | add 5 | 2x^{2} + 5 | square root | √2x^{2} + 5 |
h | g |
Exercise 3: Evaluate each of the following in their lowest terms.
Solution: For the functions: ƒ(x) = 2x + 1 and g(x) = x − 3 the function ƒg is:
The flow diagram is:
x | subtract 3 | x − 3 | multiply by 2 | 2(x − 3) | add 1 | 2(x − 3) + 1 |
g | f |
The function fg can also be determined as follows. The two functions can be written as ƒ(z) = 2z + 1 and g(x) = x − 3. Then, by substituting z = g(x) into ƒ(z) = 2z + 1:
ƒg(x) | = | ƒ(g(x)) |
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= | 2g(x) + 1 | |
= | 2(x − 3) + 1 | |
= | 2x − 5 |
Solution: For the functions: ƒ : x 2x − 1 and g : x x^{2} the function ƒg is:
The flow diagram is:
x | square | x^{2} | multiply by 2 | 2x^{2} | subtract 1 | 2x^{2} − 1 |
g | f |
The composition may also be determined by writingƒ(z) = 2z − 1 and g(x) = x^{2} and by substituting z = g(x) obtaining:
ƒg(x) | = | ƒ(g(x)) |
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= | 2g(x) + 1 | |
= | 2(x^{2}) − 1 | |
= | 2x^{2} − 5 |
Solution: For the functions: ƒ : x x^{2} and g : x 2x − 1 the function ƒg is:
The flow diagram is:
x | multiply by 2 | 2x | subtract 1 | 2x − 1 | square | (2x − 1)^{2} |
g | f |
Alternatively, writingƒ(z) = z^{2} and g(x) = 2x − 1, then substituting for z = g(x):
ƒg(x) | = | ƒ(g(x)) |
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= | (g(x))^{2}_{ } | |
= | (2x − 1)^{2}_{ } |
Solution: For the functions: ƒ : x x + 3 and g : x x − 3 the function ƒg is:
The flow diagram is:
x | subtract 3 | x − 3 | add 3 | (x − 3) + 3 |
g | f |
The function which maps x to x is called the identity function. The identity function ƒ : x x does not change the value of x.
Solution: For the functions: ƒ : x x3 − 2 andg : x 3x^{2} the function ƒg is:
The flow diagram is:
x | square | x^{2} | multiply by 3 | 3x^{2} | divide by 3 | x^{2} | subtract 2 | x^{2} − 2 |
g | f |
Writingƒ(x) = z3 and g(x) = 3x^{2}, and substituting z = g(x):
ƒg(x) | = | ƒ(g(x)) |
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= | g(x)3 − 2 | |
= | 3x^{2}3 − 2 | |
= | x^{2} − 2_{ } |
Solution: For the functions: ƒ : x 3x^{2} and g : x x3 − 2 the function ƒg is:
The flow diagram is:
x | divide by 3 | x3 | subtract 2 | x3 − 2 | square | (x3 − 2)^{2} | multiply by 3 | 3(x3 − 2)^{2} |
g | f |
Alternatively, the functions are ƒ(z) = 3z^{2} and g(x) = x3 − 2. Thus:
ƒg(x) | = | ƒ(g(x)) |
---|---|---|
= | 3(g(x))^{2} | |
= | 3(x3 − 2)^{2} |
Click on questions to reveal their solutions
The functions in question (b) above are reversed in question (c). The results show that, in general, reversing the order of two functions changes the composite function. In simpler terms, if h and k are two functions then, in general, hk ≠ kh.
Quiz 2: Two functions are given as: ƒ : x 2x^{2} − 1 and g : x x − 3. Which of the following is a solution to ƒ(g(x)) = 7?
Explanation: For the functions ƒ : x 2x^{2} − 1 and g : x x − 3, the composite function is ƒg(x) = 2(x − 3)^{2} − 1:
2(x − 3)^{2} − 1_{ } | = | 7 |
2(x − 3)^{2}_{ } | = | 7 + 1 = 8 |
(x − 3)^{2}_{ } | = | 4 |
(x − 3) | = | ±2 |
x | = | 3 ± 2 |
- so that x = 5 and x = 1 are both solutions.
If a function ƒ maps m to n, then the inverse function, written as ƒ^{−1}, maps n to m.
Example 5: Find the inverse of the function: h : x 4x − 32
Solution: First draw a flow diagram for the function:
Now draw a flow diagram, starting from the right, with each operation replaced by its inverse:
The inverse of h : x 4x − 32 is thus h^{−1} : x 2x + 34
If a function ƒ has an inverse ƒ^{−1} then the composite function ƒƒ^{−1} is the identity function which was mentioned in Exercise 3(d), i.e. ƒƒ^{−1} : x x. It is also true that ƒ^{−1}ƒ : x x.
Example 6: For the function h in Example 5, show that the composite function hh^{−1} is the identity function.
Solution: First note that h^{−1} : x 2x + 34. For the composition hh^{−1}, therefore, this must be operated on by the function h, i.e. in the first flow diagram of Example 5 the input on the right hand side must be 2x + 34.
x | h^{−1}(x) | 2x + 34 | multiply by 4 | 2x + 3 | subtract 3 | 2x | divide by 2 | x |
h(x) = 4x − 32 |
Quiz 3: Two functions are given as: ƒ : x 1 2 x and g : x 3x + 16. If h = ƒg, which of the following is h^{−1}?
Explanation: The function h = ƒg is h : x 12(3x + 16). This is the function obtained by simplifying the function of Exercise 4(e). The inverse of the function is thus:
If three functions are given as: ƒ : x x^{2}, g : x 4x, h : x x + 5, choose the correct options for the following