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The graph above is concave up on the intervals: $[-5,0]$ and $[0,5]$. My question is: Is the graph concave up on this interval $[-5,5]$ ? In other words: Since $x=0$ is a corner, does that effect the concavity of the interval $[-5,5]$ ?

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2 Answers

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A function is concave up (also called convex) on an interval $I\subset\mathbb R$ if $$f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$$ for all $a,b\in I$ and for all $t\in [0,1]$. Geometrically, this simply means that the line connecting two points $(a,f(a))$ and $(b,f(b))$ does not dip below the graph of the function over the interval with endpoints $a$ and $b$.

Your function does not satisfy this condition for $a=-1$, $b=1$, and any $t\in(0,1)$. Therefore, it is not concave up on $[-1,1]$ (or any interval whose endpoints lie on opposite sides of the origin).

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In order for an attribute like this to be true over an interval, the thing it's measuring must also exist over that interval. Since at $x=0$, this function doesn't have a second derivative, it can't be considered to be concave up (or down) on any interval that includes $0$ in its interior.

Even worse, if we were to construct a series of 2-smooth functions that increasingly approximate the function around 0, the resulting derivative at 0 would be increasingly negative, and thus definitely not concave up.

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