Determine whether the following functions with their specified domain and range is injective, surjective, and bijective. If you determine that a given function with its specified domain and range is injective, surjective, or bijective, you do not need to prove it. On the other hand, if you determine that a given function with its specified domain and range fails to be injective, surjective, or bijective, you MUST provide a counter example.

a. f: R → R defined by f(x) = x^4 + 1.
b. f: R (1,0) defined by f(x) = x^4 + 1
c. f: R + R defined by f(x) = 2^x.
d. f: R (1,0) defined by f(x) = 2^x + 1.

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]

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