Define a function f : R → R by the formula f (x) = 3x − 5.
(a) Prove that f is one-to-one.
(b) Prove that f is onto.

2 thoughts on “Define a function f : R → R by the formula f (x) = 3x − 5.<br /> (a) Prove that f is one-to-one.<br /> (b) Prove that f is onto.”

1. Solution−

   Given that,

   [tex]\rm :\longmapsto\:f(x) = 3x – 5:⟼f(x)=3x−5 [/tex]

   One – one

   A function f(x) defined from A to B, is said to be one – one if corresponds to one element of A, there is one image in B or no two elements of A have same image in B.

   Let us consider two elements,

   [tex]\rm :\longmapsto\:x, \: y \: \in \: R \: such \: that \: f(x) = f(y) [/tex]

   [tex]\rm :\longmapsto\:3x – 5 = 3y – 5 [/tex]

   [tex]\rm :\longmapsto\:3x = 3y[/tex]

   [tex]\bf\implies \:x = y[/tex]

   Hence,

   [tex]\bf\implies \:f(x) \: is \: one \: – \: one[/tex]

   Onto :-

   A function f(x) defined from A to B is called onto iff every element of B has a pre – image in A.

   Let if possible there exist an element y belongs to B, such that

   [tex] \rm :\longmapsto\:y = f(x)[/tex]

   [tex]\rm :\longmapsto\:y = 3x – 5[/tex]

   [tex]\rm :\longmapsto\:y + 5 = 3x[/tex]

   [tex]\rm :\longmapsto\:x = \dfrac{y + 5}{3} [/tex]

   [tex]\rm :\longmapsto\:As \: y \: \in \: R [/tex]

   So,

   [tex]\rm :\longmapsto\:\dfrac{y + 5}{3} \: \in \: R [/tex]

   [tex]\bf\implies \:x \: \in \: R[/tex]

   Hence,

   [tex]\bf\implies \:f(x) \: is \: onto.[/tex]

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2. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given that,

   [tex]\rm :\longmapsto\:f(x) = 3x – 5[/tex]

   One – one

   A function f(x) defined from A to B, is said to be one – one if corresponds to one element of A, there is one image in B or no two elements of A have same image in B.

   Let us consider two elements,

   [tex]\rm :\longmapsto\:x, \: y \: \in \: R \: such \: that \: f(x) = f(y)[/tex]

   [tex]\rm :\longmapsto\:3x – 5 = 3y – 5[/tex]

   [tex]\rm :\longmapsto\:3x = 3y [/tex]

   [tex]\bf\implies \:x = y[/tex]

   Hence,

   [tex]\bf\implies \:f(x) \: is \: one \: – \: one.[/tex]

   Onto :-

   A function f(x) defined from A to B is called onto iff every element of B has a pre – image in A.

   Let if possible there exist an element y belongs to B, such that

   [tex]\rm :\longmapsto\:y = f(x)[/tex]

   [tex]\rm :\longmapsto\:y = 3x – 5[/tex]

   [tex]\rm :\longmapsto\:y + 5 = 3x[/tex]

   [tex]\rm :\longmapsto\:x = \dfrac{y + 5}{3} [/tex]

   [tex]\rm :\longmapsto\:As \: y \: \in \: R \: [/tex]

   So,

   [tex]\rm :\longmapsto\:\dfrac{y + 5}{3} \: \in \: R[/tex]

   [tex]\bf\implies \:x \: \in \: R[/tex]

   Hence,

   [tex]\bf\implies \:f(x) \: is \: onto.[/tex]

   Additional Information :-

   1. Let us consider two sets A and B such that n(A) = n and n(B) = m and n(B), then number of one – one functions from A to B is given by

   [tex]\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:Number \: of \: one – one \: {f}^{n} -\begin{cases} &\sf{0 \: \: if \: n > m} \\ &\sf{P(m,n) \: if \: n \leqslant m} \end{cases}\end{gathered}\end{gathered}[/tex]

   2. One – one function is also called injective function.

   3. Onto function is also called subjective function.

   4. If function is both one – one and onto, then function is called bijective function.

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