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Detarmine whether each of these functions is a bijection from R to R

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The function f is a one-to-one correspondence, or a bijection, if it is both one-to-one and onto. We also say that such a function is bijective.

a) f(x)=-3x+4

Let x_1,x_2 \in\R and let us assume f(x_1)=f(x_2)

So,


f(x_1)=f(x_2)\\ \Rightarrow-3x_1+4=-3x_2+4\\ \Rightarrow x _1=x_2

Hence, we have f(x_1)=f(x_2) implies x_1=x_2 .

So, f is one-one (injective).

Also we know


-\infin<x<\infin\\ \Rightarrow -\infin<-3x<\infin\\ \Rightarrow -\infin<-3x+4<\infin\\ \Rightarrow -\infin<f(x)<\infin


So, we clearly observe the Co-Domain is the same as the Range, so f(x) is surjective.

Therefore f(x)=-3x+4 is a bijection.


b) f(x)=-3x^2+7


f(-1)=-3(-1)^2+7=4

f(1)=-3(1)^2+7=4

We have f(-1)=4=f(1), but -1\not=4.

So, f is not one-one (injective).

Therefore f(x)=-3x^2+7 is not a bijection.


c) f(x)=\dfrac{x+1}{x+2}

The function f(x) is not defined at x=-2.

Therefore f(x)=\dfrac{x+1}{x+2} is not a bijection from \R to \R.

d) f(x)=x^5+1

Let x_1,x_2 \in\R and let us assume f(x_1)=f(x_2)

So,


f(x_1)=f(x_2), x_1\in\R, x_2\in \R\\ \Rightarrow x_1^5+1=x_2^5+1\\ \Rightarrow x _1=x_2

Hence, we have f(x_1)=f(x_2) implies x_1=x_2 .

So, f is one-one (injective).

Also we know


-\infin<x<\infin\\ \Rightarrow -\infin<x^5<\infin\\ \Rightarrow -\infin<x^5+1<\infin\\ \Rightarrow -\infin<f(x)<\infin


So, we clearly observe the Co-Domain is the same as the Range, so f(x) is surjective.

Therefore f(x)=x^5+1 is a bijection.

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Question ID: mtid-5-stid-8-sqid-1202-qpid-940