Let f(x)=⌊x/2⌋. We learned that the floor and the ceiling functions are NOT invertible, but we also learned about the set of preimages of any value in the Range, the set of images. Keeping that in mind, give your answer in interval notation if necessary.

Find f−1({4}).

Respuesta :

Answer:

Value of [tex]f^{-1}(\{4\})[/tex] is 8.

Step-by-step explanation:

Given function,

[tex]f(x)=\big[\frac{x}{2}\big][/tex]

we have to find : [tex]f^{-1}(\{4\})[/tex]

It is known that,

[tex][x][/tex]=the greatest integer <=x

Then,

[tex]f(x)=\big[\frac{x}{2}\big][/tex]

[tex]\implies x=f^{-1}(\big[\frac{x}{2}\big])[/tex]

Taking x=8 we get,

[tex]f^{-1}(\big[\frac{x}{2}\big])=x[/tex]

[tex]\implies f^{-1}(\big[\frac{8}{2}\big])=8[/tex]

[tex]\implies f^{-1}([4])=8[/tex]

[tex]\implies f^{-1}(\{4\})=8[/tex]                                          ( Since [4]=4 )

Hence the value of [tex]f^{-1}(\{4\})[/tex] is 8.