find the inverse function

[jokerman]

1. Find the formula for the inverse function for w(x) = 7x^3 - 6 / 5

i substituted y for w(x) then switched around the x and the y which is how you get the inverse function i was told and when i finished i got an answer of w^-1(x) = cubeROOT 5/7x - 6/7

"5/7x - 6/7" is all inside of a cube root

2. Find the formula for the inverse function for h(x) = squareROOT x+4

for this one i followed the same way i did the first and got an answer of h^-1(x) = x^2 - 4

**my book keeps telling me to verify the inverse function on these but i didnt see anywhere that it showed how to do so, if somebody could show me how to verify these that would be nice.**

Im new to this kind of math so i kinda need someone to check my work and see if im doing it right, if im not please show me the way to do it right. thanks

 

[wjm11]

1. Find the formula for the inverse function for w(x) = 7x^3 - 6 / 5

i substituted y for w(x) then switched around the x and the y which is how you get the inverse function i was told and when i finished i got an answer of w^-1(x) = cubeROOT 5/7x - 6/7

"5/7x - 6/7" is all inside of a cube root

2. Find the formula for the inverse function for h(x) = squareROOT x+4

for this one i followed the same way i did the first and got an answer of h^-1(x) = x^2 - 4

**my book keeps telling me to verify the inverse function on these but i didnt see anywhere that it showed how to do so, if somebody could show me how to verify these that would be nice.**

Im new to this kind of math so i kinda need someone to check my work and see if im doing it right, if im not please show me the way to do it right. Thanks

You’re on the right track, but watch your math:

y = 7x^3 –6/5
Switch variables:
x = 7y^3 – 6/5
Add 6/5 to both sides:
x + 6/5 = 7y^3
Divide both sides by 7:
(x + 6/5)/7 = y^3
Take the cube root of both sides:
[(x + 6/5)/7]^(1/3) = y = w^-1(x)

To verify/demonstrate that two functions (say f(x) and g(x)) are inverses of each other, the following must be true:
Check using composite functions:
f(g(x)) = g(f(x)) = x

Additionally, a graphical check may be helpful; inverse functions will be symmetric about the line y = x.

 

[jokerman]

i guess i should have clarified that all of "7x^3 - 6" is divided by 5, not just the 6.

 

[jokerman]

and can somebody show me how to verify the first one is an inverse function. examples work way better then definitions for me.

 

[wjm11]

i guess i should have clarified that all of "7x^3 - 6" is divided by 5, not just the 6.
and can somebody show me how to verify the first one is an inverse function. examples work way better then definitions for me.

Group your expressions using parentheses as necessary.
W(x) = y = (7x^3 – 6)/5
Switch variables:
x = (7y^3 – 6)/5
Multiply both sides by 5:
5x = (7y^3 – 6)
Add 6 to both sides:
5x + 6 = 7y^3
Divide both sides by 7:
(5x + 6)/7 = y^3
Take the cube root of both sides:
[(5x + 6)/7]^(1/3) = y = w^-1(x)

Check using composite functions:
f(g(x)) = g(f(x)) = x

Example: f(x) = x^3; g(x) = x^(1/3)

f(g(x)) = f(x^(1/3)) = (x^(1/3))^3 = x
g(f(x)) = g(x^3) = (x^3)^(1/3) = x
Therefore f(x) and g(x) are inverses of each other.

Try graphing both these functions and see what they look like. Also, graph each of your functions and see what they look like.

 

[jokerman]

my teacher showed us something in class that was kinda like a trap where x^2 and squareROOT x are not inverses of eachother because one turns out to be x and the other is the absolute value of x. does the second one i asked turn out similarly or are they really inverses of eachother. if somebody could walk me through the verifying of the second one i asked that would be great. thanks

 

[tkhunny]

Of course f(x) = x^2 and g(x) = sqrt(x) are inverses, at least in some general sense. Are they "Inverse Functions"? Well, that's a different question. Let's ask the definition:

f(g(x)) = (sqrt(x))^2 = x -- That worked.
g(f(x)) = sqrt(x^2) = |x| -- That didn't.

 

[stapel]

my teacher showed us something in class that was kinda like a trap where x^2 and squareROOT x are not inverses of eachother because one turns out to be x and the other is the absolute value of x.

The definition and reasoning has already been explained in the previous reply, but the upshot is that, if you are faced with taking the square root of a square, make sure that you are clear on the domain.

For instance, if you are given f(x) = x², x < 0, then the inverse function is not g(x) = sqrt[x], but rather h(x) = -sqrt[x].

Check the graphs, if you're not sure about this! :wink:

Eliz.