Analyzing One-to-One Functions and Parametric Curves

[System]

Homework Statement

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) :

x(t)=2sin^2(t) and y(t)=3cos^2(t) where t belongs to [0,pi/2]

Homework Equations

The Attempt at a Solution

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!

 

[Mark44]

Homework Statement

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) :

x(t)=2sin^2(t) and y(t)=3cos^2(t) where t belongs to [0,pi/2]

Homework Equations

The Attempt at a Solution

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.

That works for me.

For (2): I do not have any idea!

You know that sin²(t) + cos²(t) = 1, right? Can you work this idea into your parametric equations to get one equation that involves only x and y (no t)?

 

[System]

well,
I would say:
(x/2)+(y/3)=sin^2(t)+cos^2(t)=1
so
(x/2)+(y/3)=1
which is ellipse
ok
then?

 

[Mark44]

No, x/2 + y/3 = 1 is NOT an ellipse. Also, keep in mind that 0 <= t <= pi/2.

 

[System]

ohhh
Sorry
its a line
so when i substitute the smallest value for t i will get the start of the line
and when i substitute the biggest for t i will get the end of the line
so its TRUE

 

[System]

by the way.
the first statement is true.
its one-to-one function.

 

[Mark44]

by the way.
the first statement is true.
its one-to-one function.

Right. I stand corrected. I should have drawn the graph, because then it's obvious.