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

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To find the inverse of the function y = x^3 + x + 4, the initial step involves switching the variables to x = y^3 + y + 4. Isolating y proves challenging, as it leads to a cubic equation y^3 + y + 3 = 0. Since this cubic may yield multiple y values for a single x, it indicates that the function may not have a true inverse. Numerical methods, such as the bisection method, can be employed to approximate solutions for specific values, like f^-1(1), by testing intervals to find roots. Ultimately, the complexity of solving cubic equations necessitates careful consideration of the function's behavior to determine invertibility.
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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.
 
The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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