Finding inverse functions when there is X^2+x

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Homework Help Overview

The discussion revolves around finding the inverse of the function f(x) = 2x + x^2, which involves solving a quadratic equation. Participants are exploring the implications of the function's structure on its invertibility.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss how to manipulate the equation x^2 + 2x = y to isolate x. There are attempts to rewrite the equation into a standard quadratic form and solve it. Questions arise regarding the necessity of including both positive and negative roots in the solution.

Discussion Status

The discussion is active, with participants providing feedback on each other's attempts. Some guidance has been offered regarding the need to consider the function's domain for invertibility, and there is acknowledgment of mistakes in the proposed solutions.

Contextual Notes

There is a note that the function y = x^2 + 2x does not represent a one-to-one function, which raises questions about the existence of an inverse unless domain restrictions are applied.

applestrudle
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Homework Statement



f(x)= 2x + x^2

Homework Equations





The Attempt at a Solution



I don't know how to make x the subject
 
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applestrudle said:

Homework Statement



f(x)= 2x + x^2

Homework Equations





The Attempt at a Solution



I don't know how to make x the subject

You need to solve for x in the equation x^2 + 2x = y. This is a simple quadratic equation.
 
Ray Vickson said:
You need to solve for x in the equation x^2 + 2x = y. This is a simple quadratic equation.

Okay thank you!

I've got it:

X^2 +2x = y

add one so you get a quadratic that has a "squared"

x^2 +2x +1 = y + 1

solve quadratic:

(x+1)^2 = y + 1

x = (y+1)^-1/2 -1

f-1(x) = (x+1)^-1/2 -1

:)
 
Last edited:
applestrudle said:
Okay thank you!

I've got it:

X^2 +2x = y

add one so you get a quadratic that has a "squared"

x^2 +2x +1 = y + 1

solve quadratic:

(x+1)^2 = y + 1

x = (y+1)^-1/2 -1
No.
Starting from two lines above, you want to solve for x.
When you do this, don't forget that you need ± in there somewhere.
applestrudle said:
f-1(x) = (x+1)^-1/2 -1

:)
Note that the equation y = x2 + 2x does not give y as a 1-to-1 function of x, so there is no inverse function, unless you place restrictions on the domain.
 
applestrudle said:
f-1(x) = (x+1)^-1/2 -1

:)
You have raised to the power -1/2. Where did the negative sign come from?
 
oay said:
You have raised to the power -1/2. Where did the negative sign come from?

Sorry, mistake when I was typing! Thank you for pointing it out
 

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