# Homework Help: Inverse Function

1. Apr 7, 2009

### SETHOSCOTT

1. The problem statement, all variables and given/known data

Find the inverse function.
[Find (f^-1)(x)]
f(x) = (-2x^5) + 1/3

2. Relevant equations

Find the inverse function.
[Find (f^-1)(x)]
f(x)=2-2x^2

3. The attempt at a solution

(-2x^5) + 1/3
(-1/2x^1/5) + 3
Answer is different, how would I get to (-1/2x + 1/6)^1/5?

Last edited: Apr 7, 2009
2. Apr 7, 2009

### tiny-tim

uhh? if x → y is a function, then the inverse is y → x.

so if y = -2x5 + 1/3, what is x ?

3. Apr 7, 2009

### SETHOSCOTT

Yea, not for this, just find functions.

4. Apr 7, 2009

### SETHOSCOTT

Is the answer of the referred problem even correct?

5. Apr 7, 2009

### Staff: Mentor

If you have y = f(x) = -2x5 + 1/3, then the inverse is x = f-1(y).
From your equation y = -2x5 + 1/3, solve for x in terms of y. That will be your inverse, as a function of y. To write the inverse as a function of x, just switch the variables (i.e., change y to x).

6. Apr 7, 2009

### Staff: Mentor

Not even close. What you did is not how you find the inverse function. See tiny-tim's and my posts.

7. Apr 7, 2009

### SETHOSCOTT

Inverse of x=y does not = y=x.
-y=-x is the inverse as it is perpendicular.
Go graph it.

8. Apr 7, 2009

### SETHOSCOTT

It's the answer in the book, the reference has less than a 1/1,000 chance of being wrong.

9. Apr 7, 2009

### tiny-tim

Like this

10. Apr 7, 2009

### SETHOSCOTT

I still want to know how, though.

11. Apr 7, 2009

### SETHOSCOTT

There is no reference to what x is.

12. Apr 7, 2009

### skullers_ab

I am confused with all the confusion here.

@SETHOSCOTT
tiny-tim and Mark44 have given the perfect cues to the answer.

It's a simple algebraic manipulation of the equation. Interchange x and y and make y the subject of the equation to get the inverse function.

What's left to clear up?

13. Apr 7, 2009

### Staff: Mentor

The function y = x is its own inverse. Perpendicularity has nothing to do with inverses.

-y = -x is in fact equivalent to y = x. The graphs of these two equations are identical.

14. Apr 8, 2009

### SETHOSCOTT

I'm not going to even say anything... nevermind...

15. Apr 9, 2009

### SETHOSCOTT

PWND
y=-2x5 + (1/3)
- (1/3) - (1/3)
y - (1/3) = -2x5
* -.5 * -.5
(1/6) - (y/2)= x5
[(1/6) - (y/2)]1/5 = (x5)1/5
[(1/6) - (y/2)]1/5 = x

16. Apr 9, 2009

### SETHOSCOTT

Don't kid yourself... har-har.

17. Apr 9, 2009

### tiny-tim

That's better!

But you don't need all those intermediate steps (some of which I don't really understand) …

it's ok if you just say

y = -2x5 + (1/3),

so 2x5 = (1/3) - y

so x = (1/6 - y/2)1/5

(and so the inverse is f-1(x) = … ? )

18. Apr 9, 2009

### SETHOSCOTT

Wait so you KNEW this?

19. Apr 9, 2009

### SETHOSCOTT

Well, duh, switch, ummm... x and y, but, yea.

Last edited: Apr 9, 2009
20. Apr 9, 2009

### tiny-tim

"Epiphany"

"duh" is irrelevant …

if you want full marks in the exam,

you do actually need to finish the question!

21. Apr 9, 2009

### SETHOSCOTT

I did, I put, "f-1(x)= [(1/6) - (x/2)]1/5," for my answer. TY, anyways. Off to chemistry, for metallurgy.

22. Apr 9, 2009

### SETHOSCOTT

Wait, you guys said all I needed to do was switch the variables, which is false.

23. Apr 9, 2009

### tiny-tim

where?
No, it's true … switch x and y (or x and f(x)) in the original equation, and then solve for x

24. Apr 9, 2009

### SETHOSCOTT

R1. On my assignment. I put it on my assignment.

R2. Yea, see, you didn't say anything about solving, it worked either way.

25. Apr 9, 2009

### tiny-tim

ah, you mean …
… we thought that was the answer in the book.