$\displaystyle f(x)=-x^2$ for x<0 and $\displaystyle f(x)=\sqrt{x}$ for x $\displaystyle \ge$ 0. There is a line tangent to these two graphs at two points. Find the distance between these two points.
Thanks!
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 $\displaystyle q=\dfrac{p^2}{4}$ but when i set the derviative equal to one another im getting $\displaystyle q=\dfrac{4}{p^2}$....
Let's do this one. I may have inadvertently made some typos in the last post. The result is correct, though.
$\displaystyle y=-x^{2}, \;\ y=\sqrt{x}$
Let p be a point on $\displaystyle -x^{2}$: $\displaystyle P(p,-p^{2})$
Let q be a point on $\displaystyle \sqrt{x}$:$\displaystyle Q(q,\sqrt{q})$
The slope of P is $\displaystyle -2p$
The slope of Q is $\displaystyle \frac{1}{2\sqrt{q}}$
Therefore, $\displaystyle q=\frac{1}{16p^{2}}$.......[1]
The slope from P to Q is $\displaystyle \frac{-p^{2}-\sqrt{q}}{p-q}$
This has to be the slope at P (or Q):
$\displaystyle \frac{-p^{2}-\sqrt{q}}{p-q}=-2p$
Sub in [1]:
$\displaystyle \frac{-p^{2}-\sqrt{\frac{1}{16p^{2}}}}{p-\frac{1}{16p^{2}}}+2p=0$
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.