Question

HELP! (Urgent pre-calculus question.)

Answer 1

2. means that for every x, there is eactly one y
basically, we can use the vertical line test to see if it crosses at any 2 points

first one works
2nd one works
3rd works
4th works
5th works
they are all one to one functions



3.
easy
steps
for f(x)=?
replace f(x) with y
switch x and y
solve for y
replace y with  f⁻¹(x)
so

$f(x)=(1+x)^3-2$
replace
$y=(1+x)^3-2$
switch
$x=(1+y)^3-2$
solve
$x+2=(1+y)^3$
$\sqrt[3]{x+2}=1+y$
$-1+\sqrt[3]{x+2}=$
replace
$f^{-1}(x)=-1+\sqrt[3]{x+2}$
first option



4.
domain, hmm we can't divide by 0 so domain is all real numbers except for x=0
domain is $(-\infty,0) \cup(0,\infty)$
hmm, range
as x approaches negative or positive infinity, f(x)=5
as x approaches 0 from the left, f(x) appraoches negative infintiy
as x approaches 0 from the right , f(x) appraoches positive infintiy
so range is $(-\infty,5) \cup(5,\infty)$

domain is $(-\infty,0) \cup(0,\infty)$
range is $(-\infty,5) \cup(5,\infty)$
answer is the 4th option