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Piece-wise defined functions

  • Thread starter tics
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  • #1
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Homework Statement



Please help me sketch the graph of the given piecewiswe-defined function and its transformations:
The function f(t)={t+4 ;-4<=t<0
={-3/2(t)+10 ; 0<=t<2
={1/2(t) ; 2<=t<=4

Homework Equations


The transformations of the function of f(t) are:
h(t)= f(t)-2 and k(t)= 3-2f(t-2)


The Attempt at a Solution



I understand that a piecewise function is defined with different
formulas. In this case, all formulas are linear or straight lines in
a form of y=mx+c.Here I'm not sure: The first formula; f(t)= t+4,
provided -4<=t<0 is a line with a of slope of 1 and an x-intercept
of t=4; but we are only supposed to sketch that part of a line that
lies to the left or at t=4.Therefore,f(4)=4+4=8 :fixed endpoint of
the ray is (4,8)

Thank
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Homework Statement



Please help me sketch the graph of the given piecewiswe-defined function and its transformations:
The function f(t)={t+4 ;-4<=t<0
={-3/2(t)+10 ; 0<=t<2
={1/2(t) ; 2<=t<=4

Homework Equations


The transformations of the function of f(t) are:
h(t)= f(t)-2 and k(t)= 3-2f(t-2)


The Attempt at a Solution



I understand that a piecewise function is defined with different
formulas. In this case, all formulas are linear or straight lines in
a form of y=mx+c.Here I'm not sure: The first formula; f(t)= t+4,
provided -4<=t<0 is a line with a of slope of 1 and an x-intercept
of t=4; but we are only supposed to sketch that part of a line that
lies to the left or at t=4.Therefore,f(4)=4+4=8 :fixed endpoint of
the ray is (4,8)

Thank
What you are saying is correct so I don't see what your difficulty is.

The graph of y= t+ 4 is a straight line. When t= -4, y= 0 and when t= 0, y= 4 so it is straight line segment with endpoints (-4, 0) and (0, 4). It includes (-4, 0) but not (0, 4).

The graph of (-3/2)t+ 10 is a straight line segment with endpoints (0, 10) and (2, 7). It includes (0, 10) but not (2, 7).

The graph of (1/2)t is a straight line segment with endpoints (1, 1) and (4, 2). It includes both (1, 1) and (4, 2).
 
  • #3
15
0
What you are saying is correct so I don't see what your difficulty is.

The graph of y= t+ 4 is a straight line. When t= -4, y= 0 and when t= 0, y= 4 so it is straight line segment with endpoints (-4, 0) and (0, 4). It includes (-4, 0) but not (0, 4).

The graph of (-3/2)t+ 10 is a straight line segment with endpoints (0, 10) and (2, 7). It includes (0, 10) but not (2, 7).

The graph of (1/2)t is a straight line segment with endpoints (1, 1) and (4, 2). It includes both (1, 1) and (4, 2).
Hi HallsofIvy, I understand now, but the last graph isn't supposed to be (2,1) because when t=2 ,y=1...I think you made a little error.

I tried to plot the f(t) using Derive6, here is my sketch below: I'm not sure with the transformations, h(t) and k(t). Please help. Thanks
 

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