Basically sin^-1(x²+2x) being not injective cause the argument of sin⁻1(x) is a parabola...Is that always a safe "assumption" or at least suspicion? Like to give another example ln(2x^2 +3x+1) is non injective cause the "inner" function is non injective (parabola again) but if you change it to ln(x^3+2 x) it is injective cause thats a cubic polynomial and those always represente injective functions.
#Is the injectivity of a function 'within' another function always down to the 'inner' function?
verbal sorrel
mighty krakenBOT
k12byda5h
verbal sorrel
yeh
limber musk
ok
yeah, for injective, you have to prove that for $x,y \in D_f$, we have $f(x) =f(y)$ implies $x=y$.
mighty krakenBOT
k12byda5h
limber musk
I mean, in general, your assumption is right
but it is a bit different in your first problem
for arctan, the domain is in [-1,1]
and the function x^2+2x is injective on that interval
wait wait
verbal sorrel
for arctan, the domain is in [-1,1]
you mean arcsin
limber musk
I think I confuse myself
limber musk
**k12byda5h**
as we mention here, you can see that $(-1.1)^2+2(-1.1) = (-0.9^2)+2(-0.9) = -0.99$
mighty krakenBOT
k12byda5h
limber musk
Therefore, x^2+2x = -0.99 for x = -0.9,-1.1
since -0.99 is in domain of arcsin, we have arcsin(x^2+2x) is equal when x = -0.9,-1.1
so they are not injective
ignore what I said above
the general answer for your prob is, it is mostly true, just be careful about the domain of function
For example, if I choose function $\sqrt{1-(x(x-1)(x+1) - 3)^2 }$
mighty krakenBOT
k12byda5h
limber musk
The polynomial $x(x-1)(x+1)$ itself is not surjective
mighty krakenBOT
k12byda5h
limber musk
but the domain of $\sqrt{1-(x-3)^2}$ is [2,4]
mighty krakenBOT
k12byda5h
limber musk
and $x(x-1)(x+1)$ is injective on the interval that its range is in [2,4]
mighty krakenBOT
k12byda5h
