• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

confusing graphs question (1 Viewer)

bally24

Member
Joined
Mar 13, 2006
Messages
53
Gender
Female
HSC
2006
in fitz exercise 30a it asks you to graph y=sqrt ((cosx - 1)<sup>2</sup>) Can't you simplify this to y=cosx - 1 and graph that? But when I graphed it on my graphing program, I was wrong. I know Im missing something but I'm starting to get really confused about it - could someone clarify this for me? (my half yearly's tomorrow :( :worried:)

Heres the graphs of y=sqrt ((cosx - 1)<sup>2</sup>), y=((cosx - 1)<sup>2</sup>) and y=cosx - 1.
 
P

pLuvia

Guest
y=sqrt[(cosx-1)2] does not equal to y=cosx-1
y=sqrt[x2]=y=|x| that means your graph should be y=|cosx-1|

Hope that helped
 

Riviet

.
Joined
Oct 11, 2005
Messages
5,593
Gender
Undisclosed
HSC
N/A
Remember that sqrt[f(x)] is always positive and is undefined when f(x)<0. Using what pLuvia said that sqrt{f(x)2}=|f(x)|, the graph of y=|f(x)| is basically the graph of y=f(x), except with the parts of the graph below the x-axis reflected in the x-axis (since f(x) < / 0).
 
Last edited:

bally24

Member
Joined
Mar 13, 2006
Messages
53
Gender
Female
HSC
2006
oh right, i tmakes more sense when i think about it in terms of y=sqrt x^2 instead of all this trig, thanks for your help guys :)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top