1. Distance between two points

$f(x)=-x^2$ for x<0 and $f(x)=\sqrt{x}$ for x $\ge$ 0. There is a line tangent to these two graphs at two points. Find the distance between these two points.

Thanks!

2. This is practically the same problem I answered before. You can use the same method to find the points, then use the distance formula to find the distance between them.

3. Originally Posted by galactus
This is practically the same problem I answered before. You can use the same method to find the points, then use the distance formula to find the distance between them.
I know, but when i tries it with your method, im getting extremely large values. And the other thing is, in your original post your said that $q=\dfrac{p^2}{4}$ but when i set the derviative equal to one another im getting $q=\dfrac{4}{p^2}$....

4. Then that woukd dictate I made a typo. Sorry, just typed it backwards. I will fix it.

5. Let's do this one. I may have inadvertently made some typos in the last post. The result is correct, though.

$y=-x^{2}, \;\ y=\sqrt{x}$

Let p be a point on $-x^{2}$: $P(p,-p^{2})$

Let q be a point on $\sqrt{x}$: $Q(q,\sqrt{q})$

The slope of P is $-2p$

The slope of Q is $\frac{1}{2\sqrt{q}}$

Therefore, $q=\frac{1}{16p^{2}}$.......[1]

The slope from P to Q is $\frac{-p^{2}-\sqrt{q}}{p-q}$

This has to be the slope at P (or Q):

$\frac{-p^{2}-\sqrt{q}}{p-q}=-2p$

Sub in [1]:

$\frac{-p^{2}-\sqrt{\frac{1}{16p^{2}}}}{p-\frac{1}{16p^{2}}}+2p=0$

Now, solve for p and q. You can get points and find the distance between them.

This is how I always went about these 'line tangent to two points' problems. Maybe you can find another way.