When f" is 0 but not considered an inflection point.?

Question

When I was doing calc. I noticed that, a concaved up f" graph, touches the x axis but is not considered an inflection point. usually f"(x)=0 is an inflection point right?? I appreciate your help.

Answer 1

not necessarily.



points of inflection are the points where the concavity changes (either from concave up to concave down or vise-verse).



Points of inflection will occur only if the second derivative is zero or undefined. However, just because the second derivative is zero (or undefined) does not guarantee an inflection point. It may be a cusp.



...It's just like maximums & minimums: the first derivative being zero or undefined does not always mean a local max/min happens.

Answer 2

definite, yet you may desire to pay interest. Like Zentraed pronounced, y''(x) = 6x and, hence, y''(0)= 0. however the easy fact that y''(0) = 0 isn't sufficient to assert that the function has an inflection element at x =0. it is needed that the 2d by-product exchange that's sign there, what quite occurs with y''(x) = 6x (from unfavorable, to beneficial). So, we quite have an inflection element at (0,0). yet once you had y = x^4, then y''(x) = 12 x^2 and y''(0) = 0, yet this function does no longer have an inflection element of x =0. on your case, you additionally can verify that x = 0 is in fact an inflection element pondering the third by-product y'''(x) = 6, that's consistently beneficial. because of fact it doen't vanish at x= 0 and y'' does, we've an inflection element.

Answer 3

f''(x) = 0 is an inflection point only if f''(x) changes from negative to positive or from positive to negative at that point. If it stays positive or if it stays negative, there is no change in concavity, therefore it is not an inflection point. For example, say you when x = 2, f''(x) is 0. To check if it changes signs, substitute 1.9 for x and 2.1 for x into f''(x). If on one side it is positive and on the other side it is negative, it is an inflection point. If the sign stays the same, it is not an inflection point.

Answer 4

consider graphs of f(x)=x^3 and f(x)= x^4 In both cases f '(x) =0 and f ''(x) =0 when x=0

taking f(x) =x^3 when x<0, f '(x) >0 and when x>0 f '(x)>0 ie point of inflection as the gradient on either side of turning point have the same sign.

taking f(x)=x^4 when x<0 f '(x) is <0 and when x>0 f '(x) is >0 a minimum point.

when f' ''(x) =0 best thing to do is to go back and look at what is happening to f '(x) on either side of the turning point.

Answer 5

probably when higher order derivatives are 0///



f(x)=x has no point of inflection, f" = 0, f' = 1.



f"' = 0 and so on

Answer 6

what about trig functions?

y = sin(x) is there a point of inflection between max and min or not?

Answer 7

i think you are right