Please tell me what is the inverse of this function "f(x)=x^3 + x" thanks?

Question

I know it has an inverse because its derivitive is 3x^2 +1 so its ascending. please please help me.

Answer 1

Solving y = x^3 + x for x is equivalent to solving x^3 + x - y = 0, which has one real solution and two complex ones; you know it only has one real solution, because x^3 + x assumes any given value of y at only one point. The real solution for x would be the inverse function. There is a computationally difficult way to find this solution, using Cardano's formula, but it would be beyond the scope of any high school and most college math; I am a mechanical engineer and I've had three years of calculus between high school and college, but I never formally learned Cardano's formula. However, it is the only way to find this result. If this is for your own edification, I invite you to attempt it. If it is for homework, it is probably sufficient to set up the solution and leave it at that.

Answer 2

The inverse of the function x^3 + x = y
is found by solving

x^3 + x = y for x in terms of y.

or

x^3 + x - y = 0.

The derivative, 3 x^2 + 1 is ascending only for x > 0.

At x = 0, x^3 + x = 0.
For x > 0, x^3 + x > 0.

So we wish to find the positive value of x if
x^3 + x = y > 0.

or x^3 + x - y = 0 when y is > 0.


There is a cubic formula somewhat like the quadratic formula for solving cubic equations, but it's not useful here for a
all y values.



I would have given you the cubic formula already, but when I
evaluated it for a simple case, it gave the wrong answer.

I need to research it.

My email is [e-mail] if you wish me to followup up on it.




Anyway, usually
Much more useful are methods for approximating the nummerical value of x for a given value of y.

Write

x^3 + x = y

Divide by x to get

x^2 + 1 = y/x

x^2 = y/x - 1

x = sqrt( y/x - 1)

That is, pick a starting value for x, say, x = 1.

divide current estimate of x into 10,
subtract 1
take square root,

Repeat these three operations until the estimate for x does not change significantly.

For example,

Find inverse of x^3 + x at the point 10.

Solve x^3 + x = 10.

Start x at 1.

10/1 = 10
10 - 1 = 9
sqrt(9) = 3

10/3 = 3.33333
3.33333 - 1 = 2.333333
sqrt(2.333333) = 1.527525...

etc

After sufficient iterations, your calculations show that
x = 2 to a large number of decimal places.,

Answer 3

The conditions that either f'(x) > 0 or f'(x) < 0 over some interval has to hold.

If said interval is for x > 0 or for x < 0 then f'(x) > 0 or f'(x) < 0 holds and the inverse will exist.


fˉ¹(x) = 1/f'(x) in that interval

So, fˉ¹(x) = 1/(3x² + 1) is the inverse.

Is this what you were looking for? I am thinking this is what you were asking....hope this helps.

Answer 4

hi, to locate an inverse function you ought to swap the x and the y (the y being f(x)^-a million ) and resolve for y. as an occasion: f(x)^-a million=y y=3/x x=3/y a million/3x=y as a result, the inverse function is: f(x)^-a million=a million/3x 2. f(x)^-a million=y y=x³-9 x=y³-9 x+9=y³ ?x+9=y as a result, f(x)^-a million=?(x+9) word that as quickly as this is the inverse function, somewhat of f(x) this is f(x)^-a million! *i regulate f(x) to y to make this much less complicated on the eyes...it helped me! remember f(x) is comparable to asserting: the function of x. f(x)^-a million is comparable to asserting: the inverse function of x. good good fortune!

Answer 5

The inverse of y = x^3 + x
is the function x = y^3 + y

Th

Answer 6

Solving y=x^3+x is by no means easy!!!

There is however one way to do it by treating y as a constant and solving X^3 +x +(-y) = 0.
Then, you apply a procedure that transform the equation to another variable.
It will by messy to solve it that way and you may end up using complex numbers in your final formula.

To find this procedure, just type in "third degree polynomial roots formula" in a search engine like Google.