Smallest possible distance between two functions Find the smallest distance between the parabolas $P_1$ and $P_2$:
$$
\displaystyle P_1:y=x^2-1\\
P_2:y=2x^2
$$
That is, find the smallest possible distance between a point on $P_1$ and a point on $P_2$.

Here's what I got so far:
Let  a point on $P_1$ be ($x_1, x_1^2-1$) and a point on $P_2$ be ($x_2, 2x_2^2$).
Applying the distance formula, 
$$
\begin{align}
d &= \sqrt{(x_1-x_2)^2+(x_1^2-1-2x_2^2)^2}\\
  &= 4x^4_2+5x_2^2-4x_1^2x^2_2-2x_1x_2  
\end{align}
$$ 
After this, I don't know how to minimize this expression, any help would be appreciated.
 A: Are you sure that your expression for $d^2$ is correct? I get $(x_1)^4+4(x_2)^4-4(x_1)^2(x_2)^2-(x_1)^2+5(x_2)^2-2x_1x_2+1$, but I may have made a mistake.
Assuming your $d^2$ is correct (though I'm pretty sure it's not): consider the square of the distance formula as a function of $x_1$ and $x_2$ (which you've already got): $f(x_1, x_2) = 4(x_2)^4 + 5(x_2)^2-4(x_1)^2(x_2)^2-2x_1x_2$.
Now use $\frac{\partial f}{\partial{x_1}}$ and $\frac{\partial f}{\partial{x_2}}$ to find where the critical points of the function are (i.e., set both partial derivatives equal to $0$ and solve the system of equations). So, we have $2(8(x_2)^3-4x_2(x_1)^2+5x_2-x_1) = 0$ and $2(-x_2 - 4x_1(x_2)^2) = 0$.
Solving that system of equations will yield values of $x_1$ and $x_2$ that minimize $f(x_1, x_2)$, which minimizes $d^2$. From there you can get $d$.
Even if your expression for $d^2$ is incorrect, it's still the same idea to find the critical points.
Edit: your expression for $d^2$ is definitely incorrect. Going through this process with your expression gives the correct values ($x_1 = 0$ and $x_2 = 0$) but it also yields $d = 0$, which is obviously wrong as you can show very easily that these parabolas never meet (try to set them equal to each other and solve for $x$).
A: You want to minimize $(x-y)^2+\bigl(f(x)-g(y)\bigr)^2$.  Taking partial derivatives we get
$$2(x-y)+2\bigl(f(x)-g(y)\bigr)f'(x)=0$$
$$-2(x-y)+2\bigl(f(x)-g(y)\bigr)\bigl(-g'(y)\bigr)=0.$$
Since $f$ and $g$ don't intersect we conclude $f'(x)=g'(y)$, in our case $x=2y$.  So now we have to minimize the distance between $(2y,4y^2-1)$ and $(y,2y^2)$, namely $4y^4-3y^2+1$.  We finally arrive in $y=\pm\sqrt{3/8}$ or $y=0$.
