Find the inverse of the cubic function What is the resulting equation when $y=x^3 + 2x^2$ is reflected in the line $y=x$ ? 
I have tried and tried and am unable to come up with the answer. 
The furthest I was able to get without making any mistakes or getting confused was $x= y^3 + 2y^2$. What am I supposed to do after that step? 
 A: The general solution to the cubic equation
$$a x^3 + b x^2 + c x + d = 0$$
can be written
$$x = -\frac{1}{3 a} \left( b + \sigma C - \sigma \frac{\Delta_0}{C} \right)$$
where
$$
\Delta_0 = b^2 - 3 a c \\
\Delta_1 = 2 b^3 - 9 a b c + 27 a^2 d \\
C = \sqrt[3]{\frac{\Delta_1 \pm \sqrt{\Delta_1^2 - 4 \Delta_0^3}}{2}} \\
\sigma = 1, \frac{1}{2} \pm i \frac{\sqrt{3}}{2}
$$
Note that if you don't care about complex solutions, then you can just ignore $\sigma$.
Comparing this to our equation
$$x = y^3 + 2 y^2 \\
y^3 + 2 y^2 - x = 0$$
we see that all we need to to turn the general form at the top into our problem is substitute $y$ for $x$, $1$ for $a$, $2$ for $b$, $0$ for $c$, and $-x$ for $d$. Plugging these values into the formula and simplifying yields our solution:
$$
\Delta_0 = (2)^2 - 3 (1) (0) = 4 \\
\Delta_1 = 2 (2)^3 - 9 (1) (2) (0) + 27 (1)^2 (-x) = -27 x + 16 \\
C = \sqrt[3]{\frac{(-27 x + 16) \pm \sqrt{(-27 x + 16)^2 - 4 (4)^3}}{2}} = \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \\
y = -\frac{1}{3 (1)} \left((2) + \sigma C - \sigma \frac{(4)}{C}\right) \\
y = -\frac{2}{3} - \frac{\sigma}{3} \left(\textstyle \sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}} \displaystyle - \frac{4}{\sqrt[3]{-\frac{27}{2} x + 16 \pm \sqrt{\frac{729}{4} x^2 - 108 x}}}\right)
$$
A: Let $z = a + bi$ be a point of the curve. Let's rotate it 45 degrees clockwise ($r_1$), then we reflect over x-axis, finally we rotate 45 degrees counter clockwise ($r_2$).
$$
rot_1 = cis(-\pi/4) \\
rot_2 = \overline{rot_1} = cis(\pi/4)
$$
In order to reflect around x-axis, we just get the conjugate of the complex number.
Let $w$ be the reflected $z$. So, $w = (\overline{rot_1 \cdot z}) rot_2 = \overline{z} \cdot \overline{rot_1} \cdot rot_2 = \overline{z} \cdot cis(\pi/2) = \overline{z} \cdot i = (a-bi)i = b + ai$.
So, the map $(x, y) \rightarrow (y, x)$ do the job.
Applying it to your equation, we have that the reflected equation is $x = y^3 + 2y^2$.
A: If you plot the equation $y = x^3 + 2y^2$, you'll find that it fails the horizontal line test, and thus that it is not a one-to-one function and its inverse is not a function. So user130558's answer must be wrong since it doesn't include any $\pm$ signs.
Unless your textbook/teacher tells you otherwise, they probably expect you to simply give the result $x = y^3 + 2y^2$ that you already got.
A: Given $$ y=x^3+2*x^2 $$ solve for x:
We can use the cubic formula:
for $$0 = a x^3 + b x^2 + c x + f$$
x equals:
$$-\frac{\sqrt[3]{\sqrt{\left(27 a^2 f-27 a^2 y-9 a b c+2 b^3\right)^2+4 \left(3 a
   c-b^2\right)^3}+27 a^2 f-27 a^2 y-9 a b c+2 b^3}}{3 \sqrt[3]{2}
   a}+\frac{\sqrt[3]{2} \left(3 a c-b^2\right)}{3 a \sqrt[3]{\sqrt{\left(27 a^2 f-27
   a^2 y-9 a b c+2 b^3\right)^2+4 \left(3 a c-b^2\right)^3}+27 a^2 f-27 a^2 y-9 a b
   c+2 b^3}}-\frac{b}{3 a}
 $$
This gives us:
$$y(x) = \frac{1}{3} (-(3 \sqrt{3} \, \sqrt{27 x^2-32 x}-27 x+16)^{1/3}/2^{1/3}-(4 2^{1/3})/(3 \sqrt{3} \, \sqrt{27 x^2-32 x}-27 x+16)^{1/3}-2)$$
