Precalculus
look at every picture and answer all questions, .

1. so it has to have every x is one y and every y has one x
graph them
we see that the only ones that are one to one are the first one, the 2nd one and the 4th one

2.
solve for x and replace with f⁻¹(x) and y with x
minus 8 and cube both sides
(y-8)³=x-2
add 2
(y-8)³+2=x
replace
f⁻¹(x)=(x-8)³+2
first one

3.
this is a one to one function because theer is no vertical line that could intersect the graph more than once
ah, I see, you're program has a different one to one definition, no horizontal or vertical line can cross, lemme edit te first question again
answer is 2nd option

4. a neat trick is this:
the domain of f(x) is the range of f⁻¹(x)
the range of f(x) is the domain of f⁻¹(x)

the domain of f(x)=1/(3x+2)
hmm, can't divide by 0 so set denomenator to 0
3x+2=0
3x=-2
x=-2/3
domain is all real numbers except -2/3
(-infinity,-2/3)U(-2/3,infinity)
that's the range of f⁻¹(x)

range
ok, hmm, range is from positive initny to 0 then from 0 to positive inifnty, not including 0

so the domain of f⁻¹(x) is (-∞,0)U(0,∞)
range of f⁻¹(x) is (-∞,-2/3)U(-2/3,∞)
first option