B Second derivative and inflection points

[i_love_science]

Q: See f(t) in graph below. Does the graph of g have a point of inflection at x=4?

There is a corner at x=4, so I don't think there is a point of inflection. Does a point of inflection exist where f''(x) does not exist? The solution says there is a point of inflection, could anyone explain why? Thanks.

 

[fresh_42]

It is all about the meaning of inflection point. In my understanding, and in accordance to Wikipedia,...

In differential calculus and differential geometry, an inflection point, point of inflection, flex, or inflection (British English: inflexion) is a point on a smooth plane curve at which the curvature changes sign.

... it is a point where the curvature changes signs. However, the curvature left of $x=4$ is zero, as it is on the right of $x=4$. Curvature does not change at $x=4$. So, yes, we need differentiability.

 

[Mark44]

There is a corner at x=4, so I don't think there is a point of inflection. Does a point of inflection exist where f''(x) does not exist? The solution says there is a point of inflection, could anyone explain why?

@i_love_science, yes, there is a corner at x = 4 in the graph of f, but the question is asking about the graph of g.
Assuming that the edges in the graph of f are straight line segments, it's not difficult to find the equations of those segments, and then to calculate the integrals.

For $x \in [0, 4]$, $g(x) = \int_0^x~f(t)dt = x^2 - 4x - 4$. This is a parabola that opens upward, so the concavity is positive on this interval.

For $x \in [4, 8]$, $g(x) = \int_4^x~f(t)dt = -x^2 + 12x - 32$. This is a parabola that opens downward, so the concavity is negative on this interval. Since the concavity changes at x = 4, there is an inflection point in the graph of g.