B burt Full Member Joined Aug 1, 2019 Messages 254 Aug 4, 2019 #1 Can there be a point a on the graph f(x) such that a is an inflection point on the graph, but f"(a) does not equal zero? Last edited: Aug 4, 2019
Can there be a point a on the graph f(x) such that a is an inflection point on the graph, but f"(a) does not equal zero?
Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 16,867 Aug 4, 2019 #2 I presume you mean f"(a). Yes, it is possible that f"(a) does not exist.
B burt Full Member Joined Aug 1, 2019 Messages 254 Aug 4, 2019 #4 Also, is a graph concave up whenever it's second derivative is positive?
Dr.Peterson Elite Member Joined Nov 12, 2017 Messages 16,867 Aug 4, 2019 #5 A classic example is the piecewise function f(x) = x^2 for x >= 0, and -x^2 for x < 0. What does your textbook say about points of inflection, and about concave up functions? Quote from it, and we can discuss that.
A classic example is the piecewise function f(x) = x^2 for x >= 0, and -x^2 for x < 0. What does your textbook say about points of inflection, and about concave up functions? Quote from it, and we can discuss that.
pka Elite Member Joined Jan 29, 2005 Messages 11,978 Aug 4, 2019 #6 burt said: Also, is a graph concave up whenever it's second derivative is positive? Click to expand... Have you considered something like this? \(\displaystyle f(x)=\begin{cases}-(x+1)^2+1 &: x\le 0 \\ (x-1)^2-1 &: x>0\end{cases}\) Is \(\displaystyle x-0\) a point of inflection? Now as Prof. Peterson points out it may well depend upon the definitions in your textbook.
burt said: Also, is a graph concave up whenever it's second derivative is positive? Click to expand... Have you considered something like this? \(\displaystyle f(x)=\begin{cases}-(x+1)^2+1 &: x\le 0 \\ (x-1)^2-1 &: x>0\end{cases}\) Is \(\displaystyle x-0\) a point of inflection? Now as Prof. Peterson points out it may well depend upon the definitions in your textbook.