Prove that a function f: R → R defined by f(x) = 2x–3 is a bijective function.
Answers
Given :
f : R → R , f(x) = 2x - 3
To prove :
f(x) is a bijective function
Proof :
• Whether f(x) is one-one function :-
Let f(x1) = f(x2)
=> 2x1 - 3 = 2x2 - 3
=> 2x1 = 2x2
=> x1 = x2
Since , f(x1) = f(x2) => x1 = x2 , hence f(x) is a one-one function .
• Whether f(x) is onto function :-
Let y = f(x)
=> y = 2x - 3
=> 2x = y + 3
=> x = (y + 3)/2
For x to be real , y can be any real number .
=> Range = R
=> Range = Co-domain
Since , Range and Co-domain of the given function are equal , thus f(x) is an onto function .
Since , f(x) is one-one and onto function , thus f(x) is bijective .
Hence proved .
Answer:
Recall that F:A→B is a bijection only if F is
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , and
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .So, is f an injection?
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .So, is f an injection?Take x,y∈IR and assume that f(x)=f(y) . Therefore 2x−3=2y−3 . We can cancel out the 3 and divide by 2 , then we get x=y .
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .So, is f an injection?Take x,y∈IR and assume that f(x)=f(y) . Therefore 2x−3=2y−3 . We can cancel out the 3 and divide by 2 , then we get x=y .Is f a surjection?
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .So, is f an injection?Take x,y∈IR and assume that f(x)=f(y) . Therefore 2x−3=2y−3 . We can cancel out the 3 and divide by 2 , then we get x=y .Is f a surjection?This is a trivial case since the two sets of the function are the set of real numbers. Every x∈IR is bound to a value defined by 2x−3 .
Recall that F:A→B is a bijection only if F isinjective: F(x)=F(y)⇒x=y , andsurjective: ∀b∈B there is some a∈A such that F(a)=b .So, is f an injection?Take x,y∈IR and assume that f(x)=f(y) . Therefore 2x−3=2y−3 . We can cancel out the 3 and divide by 2 , then we get x=y .Is f a surjection?This is a trivial case since the two sets of the function are the set of real numbers. Every x∈IR is bound to a value defined by 2x−3 .Therefore: F is bijective!
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