Keys to Assignments about Mathematical Functions

6. The graph fails the vertical line test. This is because there exists multiple vertical lines that intersect the graph in more than one point. The graph of the relation therefore does not define a function.

7. Every unique x value maps onto a unique y value provided that x-8. The relation therefore defines a function.

8. It is not a function since for every value of x (input), there is more than one value for y (output).

9. The relation defines a function. This is because the graph of the relation passes the vertical line test. If vertical lines ar.e drawn along the graph, each of them will hit the graph at only one point.

10. The relation does not define a function. This is because the graph of the relation fails the vertical line test. A vertical line drawn along the graph hits the graph at more than one point.

11. Not a function. The input x value 12 has got two output y values,5 and 8.

12.Not a function since for x>0, the output y has got more than one value. For instance when x=1,y=2

13. The relation y=x3 defines a function because every input (x value) has only one output (y value).

14.y2=6x y=6x .This implies that y takes both negative and positive value for any x>0 hence the relation y2=6x is not a function.

15. The relation represented by the straight vertical line does not define a function. This is because there exists many y values along the line for a unique x value.

16. The relation {(1,-8),(3,4),(6,-4),(8,-3),(12,-8)} defines a function. Every unique x value defines defines only one y value. Domain={1,3,6,8,12} Range={-8,-4,-3,4}

17. The relation defines a function. Every element from the left is defined by only one element from the right. That is, 4 is mapped onto 5, 7 to 8 and 12 to 13.

18. 7x=4-3y 3y=4-7x y=(4-7x)/3 .A corresponding y value is obtained for every x value input in the relation. It therefore defines a function.

19. The graph representing the relation passes the vertical line test. Any vertical line drawn vertically cuts the graph at only one single point. The relation thus defines a function.

20. Domain={4,7,12} Range={5,8,13}

21. Domain={-7,-1,3,6} Range={-8,-5,0,9}

22. y=(7x-4)

Domain: (7x-4)0 7x-40 7x4 x4/7

Domain is thus [4/7,),{x|x4/7}

Range:[0,),{y|y0}

23. x=y2 y=x. x is defined over the set x0

Domain=[0,),{x|x0}

The range of the radical begins at (0,0) up to infinity. Thus

Range=[0,),{y|y0}

24. Domain={-7,-1,3,5}

Range={-7,-1,3,5}

25. Domain={4,7,12}

Range={5,8,13}

26. Let f(x)=y so that y=-x2+4

Substitute y with x and x with y such that x=-y2+4

y2=4-x which implies that y=(4-x) which is the required inverse.

27. Let f(x)=y so that y=4x+5

X=4y+5 4y=x-5 y=(x-5)/4(inverse)

28. f(x)=(x/2)-3 y=(x/2)-3

x=(y/2)-3 2x=y-6

y=2x+6

29. (2i)(3i)= (2*3)*(i*i)=6i2=0+6i2

30. (6i)(-3i)= (6*-3)*(i*i)=-18i2=0-18i2

31. The function is invertible since every x is mapped onto a particular y.

Inverse={(-14,-8),(-1,16),(11,-14)}

32. f(x)=4x+7

f(3)=4(3)+7=12+7=19