1
$\begingroup$

How do I find the intervals when the graph below is concave up and when it's concave down as well as its inflection coordinate points just by looking at the graph?

enter image description here

I'm able to eyeball it here, but I'm confused with a few parts. I'm pretty sure it concaves up in $(0, 2)\cup(5, \infty)$. I'm not sure if the region around $x=4$ should have its concavity considered because the derivative is not defined at this point. Then I suspect it concaves down in $(2, 3.5)$. Again, I'm not sure if the point after $x=4$ but just before $x=5$ should be considered an interval where it concaves down, but I doubt it.

Then I think the inflection coordinates are $(1, 4), (3, 6), (5, 5)$

$\endgroup$
3
  • $\begingroup$ $(1, 4)$ is not an inflection point. Inflection is when $f'' = 0$, i.e. the slope that was increasing stops increasing or when it was decreasing and then starts increasing. Inflection point is somewhere between $x = 1$ and $x = 3$ as slope is continuous and is increasing at $x = 1$ ($f' > 0$) and decreasing at $x = 3$($f' < 0$) So by Rolle's theorem there must be $x \in[1, 3]$ such that $f'' = 0$. $(3, 6)$ is also not an inflection point. $\endgroup$
    – ab123
    Mar 29 '20 at 7:36
  • $\begingroup$ @ab123, that is not true. $f''=0$ doesn't always mean inflection point, for example consider $x\mapsto x^4$, then the double derivative of this function is zero at origin, but that is not a point of inflection. $\endgroup$
    – Martund
    Mar 29 '20 at 8:17
  • $\begingroup$ @Martund yes, sorry about that. I should have said - If $f''$ changes sign (from positive to negative, or from negative to positive) at a point $x = c$, then there is an inflection point located at $x = c$. $\endgroup$
    – ab123
    Mar 30 '20 at 7:43
1
$\begingroup$

“Concave up” is like the graph of $x^2$ (any arc of it); “concave down” is like the graph of $-x^2$ (any arc of it).

An inflection point is where the curve has a tangent and is concave up on one side and down on the other side (not “globally”, but in some intervals).

At $(1,4)$ there is no change in concavity, and similarly at $(3,6)$. Indeed, these are a point of minimum and a point of maximum respectively.

The change in concavity happens somewhere in between $1$ and $3$ and the visual symmetry leads to guess that the inflection point is at $(2,5)$.

The point $(5,5)$ is indeed an inflection point (at least if we assume that the picture is “accurate”), because the curve is concave down before it and up past it.

What happens at $(4,3)$? This is a point where there is no tangent, so it has to be analyzed separately. The curve is concave up on either side of it, but it's nonetheless a point of minimum.

The intervals where the graph is concave up are: $[0,2]$, $[5,8]$; it is concave down on the intervals $[2,4]$, $[4,5]$.

Note that you should not say $[2,5]$, because the function is not concave down over the whole interval.

$\endgroup$
2
  • $\begingroup$ Your answers for concavity were marked correct, only issue was I needed to use parentheses instead of brackets like you did. However, I'm still not sure what all the inflection points are according to your answer $\endgroup$
    – Lex_i
    Mar 30 '20 at 5:27
  • $\begingroup$ @Lex_i They are $(2,5)$ and $(5,5)$, as mentioned in my answer. If your instructor told you that closed intervals are wrong, then they are wrong. Another instance of this common error: the function is decreasing over $[0,1]$ and $[3,4}$, increasing over $[1,3]$ and $[4,8]$; using only open intervals would be wrong. $\endgroup$
    – egreg
    Mar 30 '20 at 7:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.