A function f has domain [-1,2] and range [0,1]. The domain and range respectively of the function g defined by g(x)=1-f(x+1) is

Question

A function f has domain [-1,2] and range [0,1]. The domain and range respectively of the function g defined by g(x)=1-f(x+1) is (A) [-1,1] ;[-1,0] (B) [-2,1] ;[0,1] (C) [0,2] ;[-1,0] (D) [1,3] ;[-1,0]

Tardigrade Admin · Accepted Answer

func/sc g(x) is defined if f(x+1) is defined. Hence the domain of g is all x such that (x+1) ∈[0,2] i.e. -1 ≤(x+1) ≤ 2 ⇒ -2 ≤ x ≤ 1 (domain of g ) again f ( x +1) ∈[0,1] ∴ - f ( x +1) ∈[-1,0] 1- f ( x +1) ∈[0,1] which is the range of g. ]

Q. A function $f$ has domain $[- 1, 2]$ and range $[0, 1]$ . The domain and range respectively of the function $g$ defined by $g (x) = 1 - f (x + 1)$ is

$[- 1, 1]; [- 1, 0]$

$[- 2, 1]; [0, 1]$

$[0, 2]; [- 1, 0]$

$[1, 3]; [- 1, 0]$

Q. A function f has domain [−1,2] and range [0,1]. The domain and range respectively of the function g defined by g(x)=1−f(x+1) is

[−1,1];[−1,0]

[−2,1];[0,1]

[0,2];[−1,0]

[1,3];[−1,0]

func/sc g(x) is defined if f(x+1) is defined. Hence the domain of g is all x such that (x+1)∈[0,2] i.e. −1≤(x+1)≤2 ⇒−2≤x≤1 (domain of g ) again f(x+1)∈[0,1] ∴−f(x+1)∈[−1,0] 1−f(x+1)∈[0,1] which is the range of g. ]

Q. A function $f$ has domain $[- 1, 2]$ and range $[0, 1]$ . The domain and range respectively of the function $g$ defined by $g (x) = 1 - f (x + 1)$ is

$[- 1, 1]; [- 1, 0]$

$[- 2, 1]; [0, 1]$

$[0, 2]; [- 1, 0]$

$[1, 3]; [- 1, 0]$