Subject: Optional Mathematics
A function is a special relationship where each input has a single output. This note includes all the notes of function including its types, features and the way of representation.
A function is a special relationship where each input has a single output. Let A and B be two sets. AB is the set of all ordered pairs (a,b) such that a ∈ A and b ∈ B . Let f:AB be a non-empty subset of AB. Then f is a relation from A to B. This f is said to be a function from A to B if f associates each element of B. So, a function is a special type of relation which associates each element of the set A with one and only element of B.
Let f = {(x,y) : x ∈ A, y ∈ B} be a function from A to B. The first element x of ordered pair (x,y) is called pre-image of second element y under the function f and y is called an image of x under f. We write f(x) = y to mean y is the image of x under f and is read as f of x is y or of f of x equals y. Since y is the image of x under f, so, f(x) is an image of x under f.
Let f be a function a set A to a set B. Then, A is called domain of f and B is called codomain of f. The set of all images of the elements of an under f is called range of f. Range of f is denoted by f(A).
If A = {1,2,3}, B = {1,4,9}
f = {(1,1),(2,4),(3,9)}, then f is the function from A to B.
Here,
Domain of f = A = {1,2,,3}
Co-domain of f = B = {1,4,9}
Range of f = f(A) = {1,4,9}
Here, range of f and co-domain of f are equal sets. i.e. f(A) = B.
Again, consider a function g which is define as follows:
Here
Domain of g = A = {1,2,3}
Co-domain of g = {1,2,3,4}
In this example, a range of g i.e. g (A) is a proper subset of co-domain of g i.e. B.
Hence, a range of a function may be a proper subset of its co-domain or equal to co-domain.
So for a function f from A to B, we write f(A)≤B.
Let f be a function from A to B, then
A function can be tested in various ways. We usually test it through the definition, i.e.
a. The relation is said to be a function if all the elements of the domain must have an image in co-domain. Otherwise, it will not be the function.
The function can also be tested by using a test known as vertical line test. For this test, a vertical line is drawn in the graph at any point. If the vertical line cuts the graph at only one point it is a function and if it cuts at more than one point then it is not a function.
x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
y | 1 | 4 | 9 | 16 | 25 | 36 | 49 |
x | 0 | 1 | 2 |
f(x) | 1 | 3 | 5 |
Main features of a function
If a function f = {(1,-1),(2,1)(3,3)(4,5)}find the domain and range of f.
Solution:
Here, f = {(1,-1),(2,1)(3,3)(4,5)}
Now, Domain = set of all elements of the first set.
\(\therefore\) Domain = {1,2,3,4,5}
Range = set of all elements of the set that are mapped with first set.
\(\therefore\) Range = {-1,1,3,5}
What element in the domain has the image 8 under the function f(x) = 3x + 5?
Solution:
Here, the function f(x) = 3x + 5.
Image = 8
So, f(x) = 3x + 5
or, 8 = 3x + 5
or, 3x = 8 - 5or, 3x = 3
\(\therefore\) x = 1
So, the element 1 in the domain has the image 8 under the function.
f(x) = 3x + 5.
What will be the functional value of 4 in the function f(x) = x^{2}?
Here, f(x) = x^{2}
f(4) = ?
f(4) = (4)^{2} = 16
Let f(x) = 3x + 3 define a function in the closed interval of -2 ≤ x ≤ 2, find f(-1), f(2) and f(3).
Solution
Here, f(x) = x + 3
So f(-1) = -1 +3 = 2
f(2) = 2 + 3 = 5
Here, 3 does not belong to the closed interval of -2 ≤ x ≤ 2. So, f(3) is not defined.
The range of function f : x→2x + 3 is {1,3,5}. Find the domain of the function.
Solution:
Here, f(x) = 2x + 3
And Range = {1,3,5}
Now, f(x) = 1
or, 2x + 3 = 1
or, x = -1
Again f(x) = 3
or, 2x + 3 = 3
or, x = 0
Also, f(x) = 5
or, 2x + 3 = 5
or, x = 1
So, domain of f = {-1,0,1}
If f(x+3) = f(x) + f(3) x∈R then prove that f(0) = 0 amd f(-3) = -f(3).
Solution:
Here f(x+3) = f(x) + f(3)
Putting x = 0, we get,
f(0+3) = f(0) + f(3)
or, f(3) = f(0) + f(3)
or, f(3) - f(3) = f(0)
or, f(0) = 0 proved.
Again , putting x = -3, we get
f(-3+3) = f(-3) + f(3)
or, f(0) = f(-3) + f(3)
or, 0 = f(-3) + f(3)
or, f(-3) = -f(3) proved.
If f(x+1) = 2x - 3, find f(x) and f(2).
Solution:
Here, f(x+1) = 2x - 3
or, f(x+1) = 2(x+1) - 3 - 2
= 2(x+1) - 5
So, f(x) = 2x - 5
Now, f(2) = 2×2 - 5
= 4 - 5
= -1
If f(x) = ax + b, f(-3) = -4 and f(3) = 2,find the values of a and b.
Solution
Given that f(x) = ax + b .
f(-3) = a(-3) + b
= -3a + b
But, f(-3) = -4
So, -3a + b = -4........(i)
Also, f(-3) = a(3) + b
= 3a + b
But f(3) = 2
\(\therefore\) 3a + b = 2.........(ii)
Solving equation (i) and (ii), we get,
2b = -2
\(\therefore\) b = -1
Putting the value of b in equation (i), we get
-3a + (-1) = -4
or, -3a = -4 + 1
or, -3a = -3
a = 1
\(\therefore\) The value of a is 1 and b = -1.
What elements in the domain has the image 0 under the function f(x) = x^{2} + 5x + 6?
Solution
Let x be the element in domain which has the image 0 under the function f(x) = x^{2} + 5x + 6.
Then, f(x) = 0
or, x^{2} + 5x + 6 = 0
or, x^{2} + 3x + 2x + 6 = 0
or, x(x+3) + 2(x+3) = 0
or, (x+3) (x+2) = 0
\(\therefore\) Possible values of x are -3 and -2.
Two functions are given below:
f(x) = x^{2} + 2x - 1
g(x) = 5x + 3
Find the values of x for which f(x) = g(x).
Solution
Here, f(x) = g(x)
\(\therefore\) x^{2} + 2x + -1 = 5x + 3
or, x^{2} + 2x - 1 = 5x + 3
or, x^{2} - 3x - 4 = 0
or, x^{2} - 4x + x - 4 = 0
or, x(x-4) + x(x-4) =0
or, (x-4) (x+1) = 0
\(\therefore\) Possible values x are 4 and -1.
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