why tan(x) is not 1-1 function also tanh(x) [hyperbolic].

is it true because tan(0)=0 and tan (pie) =0 ?

Question

why tan(x) is not 1-1 function also tanh(x) [hyperbolic].

is it true because tan(0)=0 and tan (pie) =0 ?

anonymous · Answer

@Compassionate

compassionate · Answer

@dan815

anonymous · Answer

@Abhisar

anonymous · Answer

@ganeshie8

campbell_st · Answer

perhaps you could look at the domain... or even graph it.... that may help to show you where the problems lie.

kainui · Answer

Yeah you know it's not "one to one" if tan(0)=0 and tan(pi)=0. That's "two to one" isn't it?

michele_laino · Answer

please keep in mind that f(x)=tan(x) is a periodic function, by definition, whereas 
g(x)=tanh(x) is not a periodic function

perl · Answer

the language here is suggestive as Kainui pointed out. 
two x values are being sent to the same y value, that is not one to one . it is two to one.

perl · Answer

to keep going with this language , one to two would violate function-ness.
(that is, one x value being sent to two distinct y values).

perl · Answer

calling a function one to one on a domain seems intuitive than calling the function injective or surjective. sometimes the language makes things more complicated.