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
replace
switch
solve
![\sqrt[3]{x+2}=1+y](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%2B2%7D%3D1%2By)
![-1+\sqrt[3]{x+2}=](https://tex.z-dn.net/?f=-1%2B%5Csqrt%5B3%5D%7Bx%2B2%7D%3D)
replace
![f^{-1}(x)=-1+\sqrt[3]{x+2}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%3D-1%2B%5Csqrt%5B3%5D%7Bx%2B2%7D)
first option
4.
domain, hmm we can't divide by 0 so domain is all real numbers except for x=0
domain is
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
domain is
range is
answer is the 4th option
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
replace
switch
solve
replace
first option
4.
domain, hmm we can't divide by 0 so domain is all real numbers except for x=0
domain is
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
domain is
range is
answer is the 4th option