Points of Intersection of Graphed Functions

[turophile]

Here's the problem:

If two different functions have identical second derivatives, in how many points can their graphs intersect?

Here's what I have reasoned:

If two different functions have identical second derivatives, the first derivatives differ by a constant. If that constant is 0, the graphs of the first derivatives intersect in every point, in which case the graphs of the functions differ by a constant and can therefore intersect in every point (if the constant is 0), or at no point (if the constant is nonzero). If the first derivatives differ by a nonzero constant, the graphs of the first derivatives do not intersect in any point, so the graphs of the functions also do not intersect in any point. So if two different functions have identical second derivatives, the graphs of the functions can intersect in every point or at no point.

My question:

Did I get the right answer? If not, please give a hint.

 

[Deleted member 4993]

Here's the problem:

If two different functions have identical second derivatives, in how many points can their graphs intersect?

Here's what I have reasoned:

If two different functions have identical second derivatives, the first derivatives differ by a constant. If that constant is 0, the graphs of the first derivatives intersect in every point, in which case the graphs of the functions differ by a constant and can therefore intersect in every point (if the constant is 0), or at no point (if the constant is nonzero). If the first derivatives differ by a nonzero constant, the graphs of the first derivatives do not intersect in any point, so the graphs of the functions also do not intersect in any point. So if two different functions have identical second derivatives, the graphs of the functions can intersect in every point or at no point.

My question:

Did I get the right answer? If not, please give a hint.

y" - g" = 0

y' - g' = C[subtpwv4rw]1[/subtpwv4rw]

y - g = C[subtpwv4rw]1[/subtpwv4rw]x + C[subtpwv4rw]2[/subtpwv4rw]

Now you can see for y-g = 0, we can have no solution (does not intersect) or one solution (intersect at one point) depending on the values of those constant. For C[subtpwv4rw]1[/subtpwv4rw] = C[subtpwv4rw]2[/subtpwv4rw] = 0 you will have the trivial solution of y = g everywhere.

 

[mmm4444bot]

if two different functions have identical second derivatives, the graphs of the functions can intersect in every point or at no point.

f(x) = x

g(x) = -x

These graphs intersect at one point.

 

[turophile]

Thanks for the help. Following SK's example, I think the problem as stated would have a solution like this:

g'' = h'' ? g'' - h'' = 0

g' - h' = C, where C is a constant

g - h = Cx + D, where C and D are constants

So the possible points of intersection of g and h (solutions of g - h = 0) are ...

(a) No points of intersection (x ? - D/C, C ? 0)
(b) 1 point of intersection (x = - D/C, C ? 0)
(c) Intersection in all points (C and D both zero)

... depending on the values of C and D. Did I get that right?