Calculus True or False???? Help?? The graph of f(x) =1/x is concave downward for x0 and thus it has a point of inflection at x=0.

Question

Calculus True or False???? Help??

The graph of f(x) =1/x is concave downward for x<0 and concave upward for x>0 and thus it has a point of inflection at x=0.

amistre64 · Answer

how do we find a point of inflection?

amistre64 · Answer

and also,  inflection presupposes conitnuous along hte domain i believe

anonymous · Answer

I'm pretty sure point of inflection is the answers to the second derivative no?

amistre64 · Answer

yes, 2nd derivative helps define critical values for infleciton points

amistre64 · Answer

but inflection points tend to require a continuous function, at least continuous around the interval that is getting inflected

anonymous · Answer

So I've been doing this very wrong. :( Anyways, how do I know if this is true or not??

anonymous · Answer

And after this question could I post a function that needs inflection points found? Because from what you said,  I think I did it wrong.

anonymous · Answer

@mathmale @abb0t   could you help me with this?

anonymous · Answer

@eliassaab

ranga · Answer

The domain of the function f(x) = 1/x  excludes x = 0.  The function is not defined for x = 0 and therefore we cannot discuss the extrema or the point of inflection at x = 0.

ranga · Answer

If a function is continuous over an interval and it changes from concave down to concave up or vice-versa in that interval then it will have a point of inflection in that interval.

Here 1/x is concave down for x < 0 and concave up for x > 0 but at x = 0 the function is not defined and therefore x = 0 is not a point of inflection because f(x) is not continuous at x = 0.

loser66 · Answer

This site said that f"(0) >0 for x>0 and f"(0) for x<0 shows that the point 0 is inflection point although f"(0) does not exist. http://www.sosmath.com/calculus/diff/der15/der15.html at the middle of the page

ranga · Answer

@Loser66  That is a slightly different point.  
For c to be a possible point of inflection, f''(c) = 0 or f''(c) does not exist.  But f(x) should be continuous and differentiable at x = c.  In that link they tell you that the function is continuous and differentiable at all x.

Here, f(x) = 1/x.  f''(0) does not exist.  Is x = 0 an inflection point?
Since x = 0 is not in the domain of f(x), the function is not defined at x = 0 and so it cannot have a point of inflection at x = 0.

loser66 · Answer

Yes, got you, thanks @ranga

anonymous · Answer

@ranga  thank you so much