Answer:
Step-by-step explanation:
[tex]\text{Given that:}[/tex]
[tex]f: \mathbb{R} \to \mathbb{R}[/tex] [tex]\text{which is de-fined by : } f(x) = x^4 + 1[/tex]
[tex]\mathbf{f \ is \ not \ injective}[/tex]
[tex]\mathtt{counterexample:}[/tex]
[tex]Assume \ x =1, y = -1 \\ \\ f(x) = 1^4 + 1= 2 \\ \\ f(y) = (-1)^4 + 1 = 2[/tex]
[tex]Hence \ f(x) = f(y) , but \ x \ne y[/tex]
[tex]Also;[/tex] [tex]f[/tex] [tex]\text{ is not surjective.}[/tex]
[tex]Take (y )= -10 \ \varepsilon \ \mathbb{R}[/tex]
[tex]\text{But there is no pre-image x such that}[/tex] [tex]f(x) = y[/tex]
[tex](b)[/tex]
[tex]f: \mathbb{R} \to (1, \infty) \ \ \text{which is de-fined as} \ f(x) = x^4 + 1 \\ \\ \text{f is injective and surjective and; thus bijective}[/tex]
[tex](c) \\ \\ f : \mathbb{R} \to \ \mathbb{R} \text{which can be de-fined as} \ f(x) = 2^x \\ \\ \text{f is injective and f is no surjective} \\ \\ \mathtt{counterexample:} \\ \\ Assume \ 2 \ \varepsilon \ \mathbb{R} \\ \\ whereas; 2^x = -2 \\ \text{which implies that }: \implies x = \dfrac{log_x(-2)}{log 2} = not \ de'fined[/tex]
[tex](d) f : \mathbb{R} \to (1 , \infty) , \ \text{which is de'fined by} \ \ f(x) = 2^x+1 \\ \\ \text{Thus; f is injective, surjective, and bijective}[/tex]
