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