untouchablecuz
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- Mar 25, 2008
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- 2009
Sketch the function defined by:
f(x)=e^x, 0<=x<1
f(x)=f(x+1), 0<=x<=4
ok, so what i reasoned is that in the domain 0<=x<=4 it is periodic and that:
f(x+1)=e^x
.'. f(x)=e^(x-1)
however, what is troubling me is the conflicting domain, not sure how to go about graphing it
thanks
EDIT:
no worries guys, i got it
it repeats the graph of y=e^(x-1) between 0<=x<1 and so on
i.e. f(x) 0<=x<1 is identical to f(x) in 1<=x<2 and f(x) in 1<=x<2 is identical to f(x) in 2<=x<3 etc etc (except its shifted to the right)
ok
f(x)=e^x, 0<=x<1
f(x)=f(x+1), 0<=x<=4
ok, so what i reasoned is that in the domain 0<=x<=4 it is periodic and that:
f(x+1)=e^x
.'. f(x)=e^(x-1)
however, what is troubling me is the conflicting domain, not sure how to go about graphing it
thanks
EDIT:
no worries guys, i got it
it repeats the graph of y=e^(x-1) between 0<=x<1 and so on
i.e. f(x) 0<=x<1 is identical to f(x) in 1<=x<2 and f(x) in 1<=x<2 is identical to f(x) in 2<=x<3 etc etc (except its shifted to the right)
ok
Last edited: