SOLUTION: y = (x3 - 1)(1/4) A all real numbers B|x|>=1 Cx>=0 Dx>=1 E [0,1] F [-1,1] . The graph of y = SQRT(x3 - x) contains which point? A (3, 24) B(0, 1) C(1

Algebra ->  Radicals -> SOLUTION: y = (x3 - 1)(1/4) A all real numbers B|x|>=1 Cx>=0 Dx>=1 E [0,1] F [-1,1] . The graph of y = SQRT(x3 - x) contains which point? A (3, 24) B(0, 1) C(1      Log On


   

Question 803414: y = (x3 - 1)(1/4)
A all real numbers
B|x|>=1
Cx>=0
Dx>=1
E [0,1]
F [-1,1]



. The graph of y = SQRT(x3 - x) contains which point?
A (3, 24)
B(0, 1)
C(1, 1)
D(2, 6)
E (4, 2SQRT(15))
F none of the above


Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Maybe you meant y+=+%28x%5E3+-+1%29%5E%281%2F4%29
y+=+%28x%5E3+-+1%29%5E%281%2F4%29 is a function.
Its domain can be described as x%3E=1
because x%3E=1<-->x%5E3%3E=1<-->x%5E3-1%3E=0<-->y=+%28x%5E3+-+1%29%5E%281%2F4%29 exists.
The range of y+=+%28x%5E3+-+1%29%5E%281%2F4%29 can be described as y%3E=0
because y=+%28x%5E3+-+1%29%5E%281%2F4%29 exists<-->x%5E3-1%3E=0<-->y=+%28x%5E3+-+1%29%5E%281%2F4%29%3E=0

y=sqrt%28x%5E3-x%29=sqrt%28x%28x%5E2-1%29%29=sqrt%28%28x-1%29x%28x%2B1%29%29
(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%281%2A2%2A3%29=sqrt%286%29 (That eliminates point D)
For x=3 y=sqrt%282%2A3%2A4%29=sqrt%2824%29 (That eliminates point A)
For x=4 y=sqrt%283%2A4%2A5%29=sqrt%284%29sqrt%283%2A5%29=highlight%282sqrt%2815%29%29
That means that point highlight%28E%284%2C2sqrt%2815%29%29%29 is in the graph, and eliminates the answer "F none of the above".