· Show that the function f(x) = 3x – 5 is a bijective function from R to R Solution Given Function f(x) = 3x – 5 To prove The function is bijective According to the definition of the bijection, the given function should be both injective and surjectiveBijective A function f (from set A to B) is bijective if, for every y in B, there is exactly one x in A such that f(x) = y Alternatively, f is bijective if it is a onetoone correspondence between those sets, in other words both injective and surjectiveAnswer f (x) = 1x2x2 =1− 1x21 Clearly domain of f (x) is R and range is 0,1) also f (−1) = f (1) Hence this function is neither injective not surjective
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F(x)=x^2+1 is bijective
F(x)=x^2+1 is bijective-Taking the derivative of f (x) we get f' (x) = (1 x^ {2} 2x^ {2})/ (1x^ {2})^ {2}= (1x^ {2})/ (1x^ {2})^ {2} = since for all real x, 1 x^ {2} is positive and also the denominator is also positive this implies derivative is increasing also since the function is continous at x= 1 , f (x) tends infinityDemonstrating that $f(x) = x^2 1$ is bijective and calculating $f \circ f^{1}(x)$
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28/1/ · Putting f(x1) = f(x2) we have to prove x1 = x2 Since if f (x1) = f (x2) , then x1 = x2 ∴ It is oneone (injective) Check onto (surjective) f(x) = x3 Let f(x) = y , such that y ∈ Z x3 = y x = 𝑦^(1/3) Here y is an integer ie y ∈ Z Let y = 2 x = 𝑦^(1/3) = 2^(1/3) So, x is not an integer ∴ f is not onto (not surjective) Hence, function f is injective but not surjectiveAbout Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us CreatorsFunctions can be injections (onetoone functions), surjections (onto functions) or bijections (both onetoone and onto) Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true This concept allows for comparisons between cardinalities of sets, in proofs comparing the
The function f is injective or onetoone if every point in the image comes from exactly one elementinthedomainToshowafunctionisinjectiveprove x 1;x 2 2A and f„x 128/1/ · Ex 13, 6 Show that f −1, 1 → R, given by f(x) = 𝑥/(𝑥 2) is oneone Find the inverse of the function f −1, 1 → Range f (Hint For y ∈ Range · An example of a function that is not injective is f(x) = x 2 if we take as domain all real numbers If we fill in 2 and 2 both give the same output, namely 4 So x 2 is not injective and therefore also not bijective and hence it won't have an inverse A function is surjective if every possible number in the range is reached, so in our case if every real number can be reached
We have function f(x) = 3 − 4x which is defined on R Let x 1, x 2 ∈ R such that x 1 ≠ x 2 ⇒ −4x 1 ≠ −4x 2 (Multiplying both sides by 4) ⇒ 3 − 4x 1 ≠ 3 − 4x 2 (Adding 3 both sides) ⇒ f(x 1) ≠ f(x 2) This implies that each different element having different images under the function f(x)Minimum f = 463, nearly This is improved to 8sd, \displaystyle {} , using an iterative numerical method15/9/ · Thus, f is not bijective Option B f (x) = x 2 Let f(x 1) = f(x 2) ⇒ x 1 2 = x 2 2 ⇒ x 1 = x 2 ⇒ f is one one Let f(x) = y, y ∈ Z ⇒ y = x 2 ⇒ x = y – 2 ⇒ for each y ∈ Z there exists x ∈ Z (domain) such that f(x) = y ⇒ f is onto Thus, f is bijective Option C f (x) = 2x 1 Let f(x 1) = f(x 2) ⇒ 2x 1 1 = 2x 2 1 ⇒ x 1 = x 2 ⇒ f is one one Let f(x) = y, y ∈ Z ⇒ y = 2x 1 ⇒ y 1 = 2x
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F(x) = 1/x is both injective (onetoone) as well as surjective (onto) f R to R f(x)=1/x , f(y)=1/y f(x) = f(y) 1/x = 1/y x=y Therefore 1/x is one to one function that is injective Let f(x)=y 1/x = y x = 1/y which is true in Real number18/5/18 · Let A = R {3} and B = R {1} Consider the function fA→B defined by f(x)=(x2/x3) Show that f is oneone and onto and hence find f 1Let A = R {3 }, B = R {1 } If fA→ B be defined by f (x) = x 2x 3∀ x∈ A Then show that f is bijective
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To ask Unlimited Maths doubts download Doubtnut from https//googl/9WZjCW If `2f (x1)f((1x)/x)=x , then f(x)` isIf F(x)=x has no real solution then also F(F(x)=x has no real solutionFor every function f, subset X of the domain and subset Y of the codomain, X ⊂ f −1 (f(X)) and f(f −1 (Y)) ⊂ Y If f is injective, then X = f −1 (f(X)), and if f is surjective, then f(f −1 (Y)) = Y For every function h X → Y, one can define a surjection H X → h(X) x → h(x) and an injection I h(X
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The function f R → R, f(x) = 2x 1 is bijective, since for each y there is a unique x = (y − 1)/2 such that f(x) = y More generally, any linear function over the reals, f R → R , f ( x ) = ax b (where a is nonzero) is a bijectionLet A = R − (2) and B = R − (1) If f A B is a function defined by`"f(x)"=("x"1)/("x"2),` how that f is oneone and onto Hence, find f −1Answer ∴ f is an oneone function Then, y =1 The function f is onto if there x ∈ A such that f (x) = y ∴ f is onto Since f is one=one and onto then, the given function is bijective
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Find the Domain and Range f (x)=x^21 f (x) = x2 1 f ( x) = x 2 1 The domain of the expression is all real numbers except where the expression is undefined In this case, there is no real number that makes the expression undefined Interval NotationSolve for x x = (y 1) /2 Here, y is a real number When we subtract 1 from a real number and the result is divided by 2, again it is a real number For every real number of y, there is a real number x So, range of f(x) is equal to codomain It is onto function Hence it is bijective function (ii) f R > R defined by f (x) = 3 – 4x 2 Solution8/11/18 · Let f A → B be defined by f (x) = x – 2/ x – 3 ∀ x ∈ A Then show that f is bijective asked Sep 15, in Sets, Relations and Functions by Chandan01 ( 512k points)
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