Question

Which of the following is a function in their respective given domain and co-domain?

• A. $f:\{-1,0,1\}\to \{0,1,2,3\}f\left(x\right)=x^2+3$
• B. $f:\{0,1,2\}\to \{-2,-1,0,1,2\}f\left(x\right)=\sqrt{x}$
• C. $f:\{0,1,4\}\to \{-2,-1,0,1,2\}f\left(x\right)=-\sqrt{x}$
• D. $f:\{-2,-1,0,1,2\}\to \{-1,0,1\}f\left(x\right)=-|x|$

Answer 1

The correct option is C $f:\{0,1,4\}\to \{-2,-1,0,1,2\}f\left(x\right)=-\sqrt{x}$

$\{-1,0,1\}\to \{0,1,2,3\}f\left(x\right)=x^2+3$

$f(-1)=4$ doesn't belong to the codomain.
$\therefore f$ is not a function.

$\{0,1,2\}\to \{-2,-1,0,1,2\}f\left(x\right)=\sqrt{x}$

$f\left(2\right)=\sqrt{2}$ doesn't belong to the codomain.
$\therefore f$ is not a function.

$\{0,1,4\}\to \{-2,-1,0,1,2\}f\left(x\right)=-\sqrt{x}$
Now,
$f\left(0\right)=0,f\left(1\right)=-1,f\left(4\right)=-2$
The range of the function is subset of the codomain given, so it is a function by definition.

$\{-2,-1,0,1,2\}\to \{-1,0,1\}f\left(x\right)=-|x|$

$f\left(2\right)=-2$ doesn't belong to the codomain.
$\therefore f$ is not a function.