Its ok for number1.
Its true
sorry, I made mistake when i draw
its one-to-one
I need number 2 please.
1- The following function is one-to-one
f(x)= -x if x belongs to [-1,0]
f(x)=3x+2 if x belongs to [0,1]
2- The following parametric curve represents a line segment from (0,3) to (2,0) :
and where
For (1): I think its false, if we draw the function on the interval [-1,1] we will get a graph which is similar to the graph of f(x)=|x| (the v-shape). clearly there are horizontal lines intersect the curve of f in more than one point, so its not 1-1 function, so the statement is false.
For (2): I do not have any idea!
I see you already figured out (1), but try to be careful about writing intervals like [a,b] or (a,b) or (a,b] or [a,b). The f(x) you wrote above is not a function because the first part gives f(0) = 0 and the second part gives f(0) = 2.
For (2), first check start and end points.
x(0) = 0 and y(0) = 3
and .
So we can't conclude whether it's true or false yet.
Now express y in terms of x:
This is an equation of a line, so the answer to (2) is True.
Hello, Miss!
2. The following parametric curve represents a line segment from (0,3) to (2,0)
. . . . This is true.
We have: .
So we have: .
The graph would be a straight line (infinitely long).
However, ranges from 0 to
Then: .
Therefore, the graph is the line segment from (0.3) to (2,0).