Answer to Question #158130 in Discrete Mathematics for M Uzair

Question

Determine whether each of these functions is a bijection from R to R.

f(x)=2x+1

f(x)=x^2+1 

Answer 1

Here,

• $f(x)=2x+1$

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

So,

$$f(x_1)=f(x_2)\\ \Rightarrow2x_1+1=2x_2+1\\ \Rightarrow x _1=x_2$$

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

So, $f$ is one-one (injective).

Also we know

$$-\infin<x<\infin\\ \Rightarrow -\infin<2x<\infin\\ \Rightarrow -\infin<2x+1<\infin\\ \Rightarrow -\infin<f(x)<\infin$$

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

And hence $f(x)$ is bijective.

• $f(x)=x^2+1$

Here we have

$$-\infin<x<\infin\\ \Rightarrow 0\leq x^2<\infin\\ \Rightarrow 1\leq x^2+1<\infin\\ \Rightarrow 1\leq f(x)<\infin$$

So, the Co-Domain of $f$ is $\R$ , but the range of $f$ is $[1,\infin)$, so $f(x)$ is not surjective.

Hence we conclude that $f(x)$ is not bijective.