Question

) Determine whether each of these functions is a bijection from R to R.
(i) f(x) = -3x + 4
(ii) f(x) = 2x + 1
(iii) f(x) = x
2 + 1
(iv) f(x) = -3x
2 + 7
(v) f(x) = (x+1)
(x+2)

Answer 1

(i) $\forall x \in R\; \; \exist !y = -3x+4$ and $\forall y \in R \; \exist!x= \frac{4-y}{3} \Rightarrow$ bijection

(ii) $\forall x \in R\; \; \exist !y = 2x+1$ and $\forall y \in R \; \exist!x= \frac{y-1}{2} \Rightarrow$ bijection

(iii) $y = x^2+1$, $y(-1) = y(1) = 2 \Rightarrow$ is not injective, thus not bijection

(iv) $y=-3x^2+7$ , $y(-1) = y(1) = 4 \Rightarrow$ is not injective, thus not bijection

(v) $y = (x+1)(x+2)= x^2+2x+2 = (x+1)^2+1$, $y(-3) = y(1) = 5 \Rightarrow$ is not injective, thus not bijection