Help please, how do I find the inverse of the following: f(x)=(In(2-x))/(x^2+3x), h(t)= Sin^-1(t-4) and g(x)=(e^(sinx))/In(x-3)

Question

anonymous · Answer

I meant how do I find the domain of the following functions?

helder_edwin · Answer

for the first one u have to solve:
$$ \large 2-x>0 $$
and
$$ \large x^2+3x=0 $$
excluding whatever u obtain from this last.

anonymous · Answer

domain of arcsine is $[-1,1]$ so for the second one solve $-1\leq t-4\leq 1$ for $t$

anonymous · Answer

last one is trickier
domain of log is $(0,\infty)$ so you have to solve $x-3>0$ but you also have to make sure that $\ln(x-3)\neq 0$  because you cannot divide by zero

anonymous · Answer

$\ln(1)=0$ so make sure $x-3\neq 1$

anonymous · Answer

Hey Satellite, should i just use the domain of sinx as the domain for the denominator?

anonymous · Answer

which one?

anonymous · Answer

for the last one

anonymous · Answer

$$ g(x)=\frac{e^{sinx}}{\ln(x-3)}$$

anonymous · Answer

sine has no restriction, neither does $e^x$ so no worries in the numerator 
you only have to worry about the denominator

anonymous · Answer

you have to make sure that the log is defined, so $x-3>0\iff x>3$

anonymous · Answer

also you have to make sure that $\ln(x-3)\neq 0$ because you cannot divide by 0

anonymous · Answer

therefore $x-3\neq 1\iff x\neq 4$

anonymous · Answer

so your "final answer" would be $x>3,x\neq 4$ however you would like to write it,

anonymous · Answer

Thanks a lot Satellite, you really helped. Smile today!!!!

anonymous · Answer

ok i will