Question

r is a line tangent to the graphs of f=-x^2 and g=1/2 + x^2. Find r.

Answer 1

i think there is probably a snappier way to do this, but you can solve two equations
any point on once curve will be \((a, \frac{1}{2}+a^2)\) and on the other it will be \((b,-b^2)\)
the slope between those two points will be
\[\frac{\frac{1}{2}+a^2+b^2}{a-b}\]

Answer 2

on one hand that has to be equal to \(2a\) and it also must be equal to \(-2b\)

Answer 3

so you might try solving
\[\frac{\frac{1}{2}+a^2+b^2}{a-b}=2a\] and
\[\frac{\frac{1}{2}+a^2+b^2}{a-b}=-2b\] and see if you can find \(a\) and \(b\)

Answer 4

you can't do it by setting the derivatives equal, that just gives you 0

Answer 5

@satellite73 it may look snappier, but one runs into problems of uncertainty after because (maybe not in this case) there may be other points of convergence. This leads to more work and less 'snappyness' It would be best to limit one's resources to the constraints given = slope

Answer 6

slopes are the same at \(x=0\) but this only means the slope of at each curve is the same it does not tell you what line is tangent to both curves