Give the domain (that is, possible values of x):

Question

12. y= 2[SQRT(x - 1)]
A. all real numbers
B. x<1
C. -1<x<1
D. x>1
E. x>0
F. x>1/2

13. y=-[SQRT(4 - 3x)] + 6
A. -4/3<x<4/3
B. x>4/3
C. x<4/3
D. x>6
E. all real numbers
F. x>0

14. The graph of y=6 - [SQRT(4 - 3x)] contains which point?
A. (1,4)
B.(0,1)
C.(0,6)
D. (1,5)
E.(2,4)
F. none of the above

Answer 1

1) Given that y = 2[sqrt(x - 1)]

2sqrt(x - 1)>0

sqrt(x - 1)>0

x - 1>0

x >1

Therefore the answer is option ( D )

2) Given that y = -[sqrt (4 - 3x)] +6

-[sqrt(4 - 3x)]< 0

4 - 3x<0

-3x < -4

3x < 4

x<4/3

Therefore the answer is option ( C )

3) given that graph y = 6 - [sqrt(4 -3x)]

Verify the options

A) (1 , 4)

4 = 6 - [ sqrt(4 - 3(1)]

4 = 6 - [sqrt(4 -3)]

4 = 6 - sqrt (1)

4 = 6 - 1

4 = 5

(which is in correct)

B) (0 , 1)

1 = 6 - [sqrt(4 - 3(0)]

1 = 6 - [sqrt(4)]

1 = 6 - 2

1 = 4

(It is also incorrect)

C) (0 , 6)

6 =6 - [sqrt(4 - 3(0)]

6 = 6 - [sqrt(4)]

6 = 6 - 2

6 = 4

(It is also incorrect)

D) (1 , 5)

5 = 6 - [sqrt(4 - 3(1)]

5 = 6 - [sqrt(4 - 3)]

5 = 6 - [sqrt(1)]

5 = 6 - 1

5 = 5

(It is correct)

E) (2 , 4)

4 = 6 - [sqrt(4 - 3(2)]

4 = 6 - [sqrt(4 - 6)]

4 = 6 - [sqrt(-2)]

4 = 6 - 1.414

4 = 4.586

(It is also incorrect)

Therefore option ( D ) is satisfying the graph