Respuesta :

1) domain of the first is ( R ) because of its root(3) and its range is (R) because of its root again

2)and domain of the last , is [0,+infinity)

and for its range , you can draw it such as picture and will be [0,- infinity)

Ver imagen kamranT
cristianguachamin01

Answer:

1. [tex]Dom(f)=\mathbb{R} [/tex], [tex]Ran(f)=\mathbb{R} [/tex].

2. [tex]Dom(f)=\{x\in \mathbb{R} : x\geq 0 \} [/tex], [tex]Ran(f)=\{x\in \mathbb{R} : x \leq 0 \}  [/tex].

Step-by-step explanation:

1. Since the degree of the radical is an odd number, the radicand can be any real number, then [tex] x [/tex] can take any real value, so the domain of [tex] f [/tex] is the set of all real numbers, [tex] \mathbb{R} [/tex].

Now, if [tex] x\in \mathbb{R} [/tex] , then [tex] x-3 \in \mathbb{R}  [/tex], so  [tex] \sqrt[3]{x-3} \in \mathbb{R}  [/tex], and thus [tex] f(x)\in \mathbb{R} [/tex], which leads us to affirm that the range of  [tex] f [/tex]  is the set of all real numbers, [tex] \mathbb{R} [/tex].

2. Since the degree of the radical is an even number, the radicand can not be a negative number, then [tex] x [/tex] can take only nonnegaive values, so the domain of [tex] f [/tex] is the set of all nonnegative numbers, [tex] \{x\in \mathbb{R} : x\geq 0 \}  [/tex].

Now, if [tex] x\geq 0 [/tex] , then [tex]  \sqrt{x}\geq 0  [/tex] so  [tex]  -3\sqrt{x}\leq 0  [/tex], and thus [tex] f(x)\in \mathbb{R} [/tex], which leads us to affirm that the range of  [tex] f [/tex]  is the set of all nonpositive numbers, [tex] \{x\in \mathbb{R} : x \leq 0 \} [/tex].