Find the inverse of y=x^3 + x +4

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Homework Help Overview

The discussion revolves around finding the inverse of the function y = x^3 + x + 4. Participants are exploring the complexities involved in isolating y and determining whether the function has an inverse.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss switching variables and isolating y, with some expressing difficulty in this process. There is mention of the need to check for multiple y values corresponding to a single x value, which would indicate that the function does not have an inverse. Numerical methods and the use of Cardano's formula are also suggested as potential approaches.

Discussion Status

The discussion is active, with participants providing various insights and methods for approaching the problem. Some guidance has been offered regarding numerical methods and the implications of having multiple y values for a single x value, but no consensus has been reached on a specific solution or method.

Contextual Notes

There is a recognition that the original function is cubic, and participants are navigating the complexities of solving cubic equations, including the potential need for numerical solutions. The discussion also highlights the distinction between finding the inverse function and evaluating it at a specific point.

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Homework Statement




I'm helping my friend review for his exam in precalc and I can't seem to get this inverse.

Find the inverse of y=x^3 + x +4

Homework Equations





The Attempt at a Solution


I switched the variables and then tried to isolate the y but I failed.
I ended with x-4=y^3+y I don't know how to isolate the y from here how would you go about it?
 
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you would obtain x=y^3 +y +4, but unless you can find a way to produce an equivalent cubed binomial for the right side, you are stuck. If there are more than one y value for any single x value, then that is not an inverse function.
 
Given f (x) = x3 + x + 4 , find f -1 (1) accurate to 2 decimal places. Is the exact question I thought I would make it easier by just asking to find the inverse on here...
 
If you are asking for how to evaluate the inverse of f(x) at x=1, then you want to start with the inverse relationship, x= y^3 + y + 4 [I assume you still meant that the original function was a cubic], and substitute 1=x; and then find the possible values for y. If you obtain two values of y, then the inverse of f(x) is not a function.

1 = y^3 + y + 4
y^3 + y + 3 = 0
Complete the square or use quadratic formula solution;
What is y?
 
What I said in here (below) is ridiculous. You can't complete the square for a cubic; although you can look for the presence of any x which gives more than one y value.

symbolipoint said:
If you are asking for how to evaluate the inverse of f(x) at x=1, then you want to start with the inverse relationship, x= y^3 + y + 4 [I assume you still meant that the original function was a cubic], and substitute 1=x; and then find the possible values for y. If you obtain two values of y, then the inverse of f(x) is not a function.

1 = y^3 + y + 4
y^3 + y + 3 = 0
Complete the square or use quadratic formula solution;
What is y?
 
physstudent1 said:

Homework Statement




I'm helping my friend review for his exam in precalc and I can't seem to get this inverse.

Find the inverse of y=x^3 + x +4

Homework Equations





The Attempt at a Solution


I switched the variables and then tried to isolate the y but I failed.
I ended with x-4=y^3+y I don't know how to isolate the y from here how would you go about it?

physstudent1 said:
Given f (x) = x3 + x + 4 , find f -1 (1) accurate to 2 decimal places. Is the exact question I thought I would make it easier by just asking to find the inverse on here...

That's a completely different question! In order to solve the equation y3+ y+ 4= x, for any x, you would have to use Cardano's cubic formula- and that's extremely complex!

What you are asked to do here is to use some numerical method to solve the equation y3+ y+ 4= 1 or y3+ y+ 3= 0. A simple way to solve such an equation is "bijection". If y= -2, (-2)3+ (-2)+ 3= -8- 2+ 3= -7< 0. If y= -1, (-1)3+ (-1)+ 3= 1> 0 Since the value at x= -2 is negative and the value at x= -1 is positive, there must be a solution between -2 and -1. Try half way between: -3/2. Is (-3/2)3+ (-3/2)+ 3 positive or negative? Whichever it is, there must be a root between -3/2 and either -2 or -1 depending on which gave the opposite sign. Try half-way between just because it is convenient.
 

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