Answer to Question #281828 in Discrete Mathematics for Pooja

Question #281828

15.Determine whether f: R to R, defined as f (x) = āˆ’3x + 4 is a bijection. Is f invertible, and if it is, 

what is its inverse?

16.Find the inverse of f (x) =

š‘„+1

š‘„+2

, on a suitable subset of R.


1
Expert's answer
2021-12-22T09:38:46-0500

15.

Let x1,x2āˆˆRx_1, x_2 \in \R and f(x)=āˆ’3x+4.f(x) = -3x+4.

If f(x1)=f(x2)=>āˆ’3x1+4=āˆ’3x2+4f(x_1)=f(x_2)=>-3x_1+4=-3x_2+4

=>āˆ’3x1=āˆ’3x2=>x1=x2.=>-3x_1=-3x_2=>x_1=x_2.

If x1=Ģøx2=>f(x1)=Ģøf(x2).x_1\not=x_2=>f(x_1)\not=f(x_2).

Then

f(x1)=f(x2)<=>x1=x2,x1,x2āˆˆRf(x_1)=f(x_2)<=>x_1=x_2, x_1, x_2\in \R

The function f(x)=āˆ’3x+4f(x)=-3x+4 is one-to-one.


āˆ€yāˆˆR,y=āˆ’3x+4,āˆƒx=āˆ’13y+43,xāˆˆR\forall y\in\R, y=-3x+4, \exist x=-\dfrac{1}{3}y+\dfrac{4}{3}, x\in \R

The function f(x)=āˆ’3x+4f(x)=-3x+4 is onto.


The function f(x)=āˆ’3x+4f(x)=-3x+4 is one-to-one and is onto.

Therefore the function f(x)=āˆ’3x+4f(x)=-3x+4 is a bijection. Therefore the function f(x)=āˆ’3x+4f(x)=-3x+4 is invertibele


f(x)=āˆ’3x+4f(x)=-3x+4

Replace f(x)f(x) with yy


y=āˆ’3x+4y=-3x+4

Switch xx and yy


x=āˆ’3y+4x=-3y+4

Solve for yy


y=āˆ’13x+43y=-\dfrac{1}{3}x+\dfrac{4}{3}

Replace yy with fāˆ’1(x)f^{-1}(x)


fāˆ’1(x)=āˆ’13x+43f^{-1}(x)=-\dfrac{1}{3}x+\dfrac{4}{3}

16.


f(x)=x+1x+2f(x)=\dfrac{x+1}{x+2}

Domain: (āˆ’āˆž,āˆ’2)āˆŖ(āˆ’2,āˆž)(-\infin, -2)\cup (-2, \infin)

Replace f(x)f(x) with yy


y=x+1x+2y=\dfrac{x+1}{x+2}

Switch xx and yy


x=y+1y+2x=\dfrac{y+1}{y+2}

Solve for yy


xy+2x=y+1xy+2x=y+1


y=āˆ’2xāˆ’1xāˆ’1y=-\dfrac{2x-1}{x-1}

Replace yy with fāˆ’1(x)f^{-1}(x)


fāˆ’1(x)=āˆ’2xāˆ’1xāˆ’1f^{-1}(x)=-\dfrac{2x-1}{x-1}


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