Inverse Function: Solve (-2x^5) + 1/3

[SETHOSCOTT]

Homework Statement

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

Homework Equations

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

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?

 

[tiny-tim]

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

so if y = -2x⁵ + 1/3, what is x ?

 

[SETHOSCOTT]

Yea, not for this, just find functions.

 

[SETHOSCOTT]

Is the answer of the referred problem even correct?

 

[Mark44]

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).

 

[Mark44]

Is the answer of the referred problem even correct?

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

 

[SETHOSCOTT]

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

 

[SETHOSCOTT]

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

 

[tiny-tim]

how would I get to (-1/2x + 1/6)^1/5?

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

Like this …

if y = -2x⁵ + 1/3, what is x ?

 

[SETHOSCOTT]

I still want to know how, though.

 

[SETHOSCOTT]

There is no reference to what x is.

 

[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?

 

[Mark44]

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

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.

 

[SETHOSCOTT]

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

 

[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

 

[SETHOSCOTT]

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

 

[tiny-tim]

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

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) = … ? )

 

[SETHOSCOTT]

Wait so you KNEW this?

 

[SETHOSCOTT]

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

 

[tiny-tim]

"Epiphany"

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

"duh" is irrelevant …

if you want full marks in the exam,

you do actually need to finish the question! ​

 

[SETHOSCOTT]

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

 

[SETHOSCOTT]

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

 

[tiny-tim]

I did, I put, "f^-1(x)= [(1/6) - (x/2)]^1/5," for my answer.

where?

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

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

 

[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.

 

[tiny-tim]

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

ah, you mean …

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?

… we thought that was the answer in the book.

 

[SETHOSCOTT]

I guess. I think... yea, that way. a few exceptions, like utilizing the rooting overhang, instead of the powering to the 1/5, but it is equal to the answer, exactly equal.