Question 803414
Maybe you meant {{{y = (x^3 - 1)^(1/4)}}}
{{{y = (x^3 - 1)^(1/4)}}} is a function.
Its domain can be described as {{{x>=1}}}
because {{{x>=1}}}<-->{{{x^3>=1}}}<-->{{{x^3-1>=0}}}<-->{{{y= (x^3 - 1)^(1/4)}}} {{{exists}}}.
The range of {{{y = (x^3 - 1)^(1/4)}}} can be described as  {{{y>=0}}}
because {{{y= (x^3 - 1)^(1/4)}}} {{{exists}}}<-->{{{x^3-1>=0}}}<-->{{{y= (x^3 - 1)^(1/4)>=0}}}
 
{{{y=sqrt(x^3-x)=sqrt(x(x^2-1))=sqrt((x-1)x(x+1))}}}
(I transformed it because
For {{{x=0}}} and for {{{x=1}}}, {{{y=0}}}, because one of the factors in the square root is zero. (That eliminates points B and C)
For {{{x=2}}} {{{y=sqrt(1*2*3)=sqrt(6)}}} (That eliminates point D)
For {{{x=3}}} {{{y=sqrt(2*3*4)=sqrt(24)}}} (That eliminates point A)
For {{{x=4}}} {{{y=sqrt(3*4*5)=sqrt(4)sqrt(3*5)=highlight(2sqrt(15))}}}
That means that point {{{highlight(E(4,2sqrt(15)))}}} is in the graph, and eliminates the answer "F none of the above".