Functions
Let A and B be two non-empty sets. A function or a mapping f from A to be B is a rule which associates every element of A with a unique element of B and is denoted by f:A→B .
In other words, a function f from A to B is a relation satisfying
i) Every element of A is related to some element of B.
ii) No element of A is related to two different elements of B.
Domain and Range of a Function
if f:A→B is a function then A is called the domain and B is called the co-domain of f. If x∈A is associated with a unique element y∈B by the function f, then y is called the image of x under f and is denoted by y=f(x). Also, x is called the pre-image of y under f.
The range of f is the set of those elements of B which appear as the image of at least one element of A and is denoted by f(A). Thus f(A)={f(x)∈B:x∈A}. Clearly f(A) is a subset of B.
Let A = {1, 2, 3, 4} and Z be the set of integers. Define f:A→Z by f(x)=2x+3. Show that f is a function from A to B. Also find the range of f.
Solution:
Now,
Since every element of A is associated with a unique element of B, f is a function.
Therefore Range of f={5,7,9,11}
Let N be the set of natural numbers. If f:N→N is defined by f(x)=2x−1 show that f is a function and find the range of f.
Solution:
Now,
Clearly, f is a function.
Therefore, Range of f={1,3,5,7,....}
Let R be the set of real numbers. Defined f:R→Rbyf(x)=x2 for every x∈R. Show that f is a function and find the range of f.
Solution:
Here f associates every real number to its square, which is certainly a real number. Hence f is a function. Range of R is the set of all non-negative real numbers.
One-One Function or Injection
A function f:A→B is said to be one-one or injection if for all x1,x2∈A,f(x1)=f(x2)impliesx1=x2. The contrapositive of this implication is that of all x1,x2∈A,x1=x2impliesf(x1)=f(x2).
Thus a function f:A→B is said to be one-one if different elements of A have different images in B.
Let R be the set of real numbers. Defined f:R→Rbyi)f(x)=2x+3ii)f(x)=x3 for every x∈R. Prove that f is one-one.
Solution:
i) Let f(x1)=f(x2) for some x1,x2∈R
Thus for every x1,x2∈R,f(x1)=f(x2)impliesx1=x2. Therefore f is one-one.
ii) Let f(x1)=f(x2) for some x1,x2∈R
Thus for every x1,x2∈R,f(x1)=f(x2)impliesx1=x2. Therefore f is one-one.
If f:R→R is defined by f(x)=x2 for every x∈R , show that f is not one-one.
Solution:
Let
Hence f is not one-one.
For example, f(−2)=4andf(2)=4. The images of -2 and 2 are not different. Hence f is not one-one.
Onto Function or Surjection
A function f:A→B is said to be onto or surjection if for every y∈B there exist at least one element x∈A such that f(x)=y i.e., every element of the co-domain B appears as the image of at least one element of the domain A.
If f is onto, then f(A)=B.
Defined f:R→Rbyi)f(x)=2x+3ii)f(x)=x3 for every x∈R. Show that f is onto.
Solution:
i) Let y∈R then to find x∈R such that f(x)=yi.e.,2x+3=y
Solving for x we get, x=2y−3
Since y∈R,x=2y−3∈R
Hence for every y∈Rexistsx=2y−3∈Rsuch that f(2y−3)=y. Therefore f is onto.
ii) Let y∈R.
We shall show that there exists x∈Rsuch that f(x)=y.
That is x3=y.
Hence x=y31. If y∈R,theny31∈R.
Thus for every y∈R there exists y31∈R such that f(y31)=(y31)3=y.
Therefore f is onto.
If f:R→R is defined by f(x)=x2 for every x∈R then prove that f is not onto.
Solution:
Since a negative number is the square of any real number, the negative numbers do not appear as the image of any element of R. For example, −9∈R but there does not exist any x∈R such that f(x)=x2=−9. Hence f is not onto.
One-to-one function or Bijection
A function f:A→B is said to be one-to-one or bijection if it is both one-one and onto.
For example, if f:R→R is defined by i)f(x)=2x+3ii)f(x)=x3 for every x∈R then f is one-to-one function.
Inverse Image of an element
Let f:A→B be a function and y∈B. Then the inverse image of y under f denoted by f−1(y) is the set of those elements of A which have y as their image.
That is f−1(y)={x∈A:f(x)=y}.
If f:R→R is defined by f(x)=x2−3x+5 find (i)f−1(3)and(ii)f−1(15)
Solution:
i) Let
Therefore, f−1(3)={1,2}
ii) Let
Hence, f−1(15)={−2,5}
Inverse Function
If a function f:A→B is one-one and onto then the inverse of f denoted by f−1:B→A is denoted by f−1={(y,x):(x,y)∈f}
Thus if f:A→B is both one-one and onto then f−1:B→A is obtained by reversing the ordered pairs of f.
Note: f−1 exists only when f is both one-one and onto. Further f−1 is also one-one and onto.
Let Q be the set of the rational numbers. If f:Q→Q is defined by f(x)=2x−3 for every x∈Q then find f−1 if it exists.
Solution:
i) Let
Hence f is one-one.
ii) Let y∈Q. Then to find x∈Q:f(x)=y
i.e., 2x−3=y therefore x=2y+3
Whenever y is rational, x=2y+3 is also a rational. Hence there exists 2y+3∈Q such that f(2y+3)=y
Hence f is also onto. Therefore f−1:Q→Q exists.
Let x=f−1(y)⇒f(x)=y
i.e., 2x−3=ytherefore x=2y+3
Define f−1:Q→Qbyf−1(y)=2y+3 for every y∈Q.
Replacing y by x we get f−1(x)=2x+3,x∈Q. This is the required inverse function.
Composite Function or Product Function
If f:A→Band g:B→Care two functions then the composite function of f and g is denoted by gofis a function from A to C defined by (gof)(x)=g{f(x)} for every x∈A
Here (gof)(x)=g{f(x)}=g{y}=z
Note: gof=fog. That is composition of mapping is not commutative.
Let R be the set of real numbers. Define f:R→Randg:R→Rbyf(x)=3x−2andg(x)=x2+4 find (i) gof (ii) fog
Solution:
If f(x)=1−x1,x=1 find f[f{f(x)}]
Solution:
Now,
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