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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- Nov 1st 2013, 01:25 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? - Nov 1st 2013, 02:40 PMSlipEternalRe: Find the Inverse Function
Multiply both sides by 4 and let $\displaystyle y = s-t$. We want to satisfy the following system of equations: $\displaystyle s^3-t^3 = 4(x+1)$ and $\displaystyle 3st = 4$.

Replacing the $\displaystyle 4(x+1)$, and the coefficient for $\displaystyle y$, we have:

$\displaystyle s^3-t^3 = (s-t)^3+3st(s-t) = (s^3-3s^2t+3st^2-t^3) + (3s^2-3st^2)$

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

$\displaystyle s^3-t^3 = 4(x+1)$ and $\displaystyle 3st = 4$. Solving for $\displaystyle s$ in the second equation, we can plug it into the first:

$\displaystyle \left(\dfrac{4}{3t}\right)^3-t^3 = 4(x+1)$

Simplifying, we get $\displaystyle 27t^6 + 108(x+1)t^3-64 = 0$. We can now solve for $\displaystyle t^3$ by plugging it into the quadratic equation:

$\displaystyle t^3 = \dfrac{-108(x+1) \pm \sqrt{11,664(x+1)^2+6,912}}{54}$

Take the cube root of both sides, and you have $\displaystyle t$. Plug that in for $\displaystyle s = \dfrac{4}{3t}$, and that gives you $\displaystyle s$. Then, $\displaystyle y = s-t$.

(This is the general method for solving a cubic equation). - Nov 1st 2013, 02:53 PMnycmathRe: Find the Inverse Function
I'm sorry but what is the inverse?

What is f^(-1)? - Nov 1st 2013, 03:09 PMSlipEternalRe: Find the Inverse Function
$\displaystyle t = \sqrt[3]{\dfrac{-108(x+1) \pm \sqrt{11,664(x+1)^2+6,912}}{54}}$

So, $\displaystyle s = \dfrac{4}{3\sqrt[3]{\dfrac{-108(x+1) \pm \sqrt{11,664(x+1)^2+6,912}}{54}}}$

Then the inverse function is:

$\displaystyle y = \dfrac{4}{3\sqrt[3]{\dfrac{-108(x+1) \pm \sqrt{11,664(x+1)^2+6,912}}{54}}} - \sqrt[3]{\dfrac{-108(x+1) \pm \sqrt{11,664(x+1)^2+6,912}}{54}}$ - Nov 1st 2013, 06:57 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. - Nov 1st 2013, 07:06 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. - Nov 2nd 2013, 04:57 AMnycmathRe: Find the Inverse Function
You are right. It is extremely messy. I need to disregard this question.