Find the inverse of this function.

y = x^3/4 + x - 1

x = y^3/4 + y - 1

x + 1 = y^3/4 + y

x + 1 = (1/4)y[y^2 + 4]

Where do I go from here?

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- November 1st 2013, 02:25 PMnycmathFind the Inverse Function
Find the inverse of this function.

y = x^3/4 + x - 1

x = y^3/4 + y - 1

x + 1 = y^3/4 + y

x + 1 = (1/4)y[y^2 + 4]

Where do I go from here? - November 1st 2013, 03:40 PMSlipEternalRe: Find the Inverse Function
Multiply both sides by 4 and let . We want to satisfy the following system of equations: and .

Replacing the , and the coefficient for , we have:

That is obviously true. So, let's solve the system of equations:

and . Solving for in the second equation, we can plug it into the first:

Simplifying, we get . We can now solve for by plugging it into the quadratic equation:

Take the cube root of both sides, and you have . Plug that in for , and that gives you . Then, .

(This is the general method for solving a cubic equation). - November 1st 2013, 03:53 PMnycmathRe: Find the Inverse Function
I'm sorry but what is the inverse?

What is f^(-1)? - November 1st 2013, 04:09 PMSlipEternalRe: Find the Inverse Function

So,

Then the inverse function is:

- November 1st 2013, 07:57 PMnycmathRe: Find the Inverse Function
I thank you for your reply but this question is from a precalculus textbook. The answer cannot be so complicated at this level of learning.

I am to interchange x and y and then isolate y. Finally, I am to replace y with the inverse notation. After all of this, I am to evaluate the inverse by letting x be 1/2. - November 1st 2013, 08:06 PMSlipEternalRe: Find the Inverse Function
Here is Wolfram Alpha's answer: here.

It gives the complete answer. I only gave you part of the answer. In general, cubic equations are extremely messy. If that came from a precalc textbook, then my guess is that you should state that the inverse function is too messy to calculate. - November 2nd 2013, 05:57 AMnycmathRe: Find the Inverse Function
You are right. It is extremely messy. I need to disregard this question.