Question

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 1

Answer:

Step-by-step explanation:

$\text{Given that:}$

$f: \mathbb{R} \to \mathbb{R}$ $\text{which is de-fined by : } f(x) = x^4 + 1$

$\mathbf{f \ is \ not \ injective}$

$\mathtt{counterexample:}$

$Assume \ x =1, y = -1 \\ \\ f(x) = 1^4 + 1= 2 \\ \\ f(y) = (-1)^4 + 1 = 2$

$Hence \ f(x) = f(y) , but \ x \ne y$

$Also;$ $f$ $\text{ is not surjective.}$

$Take (y )= -10 \ \varepsilon \ \mathbb{R}$

$\text{But there is no pre-image x such that}$ $f(x) = y$

$(b)$

$f: \mathbb{R} \to (1, \infty) \ \ \text{which is de-fined as} \ f(x) = x^4 + 1 \\ \\ \text{f is injective and surjective and; thus bijective}$

$(c) \\ \\ f : \mathbb{R} \to \ \mathbb{R} \text{which can be de-fined as} \ f(x) = 2^x \\ \\ \text{f is injective and f is no surjective} \\ \\ \mathtt{counterexample:} \\ \\ Assume \ 2 \ \varepsilon \ \mathbb{R} \\ \\ whereas; 2^x = -2 \\ \text{which implies that }: \implies x = \dfrac{log_x(-2)}{log 2} = not \ de'fined$

$(d) f : \mathbb{R} \to (1 , \infty) , \ \text{which is de'fined by} \ \ f(x) = 2^x+1 \\ \\ \text{Thus; f is injective, surjective, and bijective}$