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I have the following

$$y = \frac{x^2}{2}-\ln x$$

$$y'= x - \frac1x$$

I learned that inflection points were found by setting the $2^{nd}$ derivative equal to $0$, however, if I do that in this case I would get $i$, and I already checked and such is not possible in this case. However when I sent the $1^{st}$ derivative equal to $0$ I get,

$$x = \pm1$$

as possible inflection points which makes more sense.

Do I have a misconception as to how to find inflection points ? Or am I missing something ?

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2  
A graphing calculator/program will show that there appear to be no inflection points. The curve is everywhere "facing upwards." Your calculation confirms that fact. – André Nicolas Feb 22 at 7:33
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"Do I have a misconception as to how to find inflection points ?": indeed. You cannot change the formulas to fit your wishes. Inflection points occur at zeroes of the second derivative, full stop. This is a necessary condition, but not sufficient. Check $y=x^4$. – Yves Daoust Feb 22 at 7:34
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I think the confusion is this: if a 0 of the first derivative is not a local maximum or local minimum then it is an inflection point. But inflection points are not all of that form. – Colin McLarty Feb 22 at 16:48
up vote 15 down vote accepted

No. Points where the first derivative vanishes are called stationary points. If the second derivative exists (as it does in this case wherever the function is defined), it is a necessary condition for a point to be an inflection point that the second derivative vanishes. Thus the fact that there are no real solutions for the equation $y''=0$ shows that the function doesn't have any inflection points.

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The definition of an inflection point is slightly imprecise; you may want to fix that. – MathematicsStudent1122 Feb 22 at 7:35
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@MathematicsStudent1122: You're right. Given how far off the suggestion about the first derivative was, it didn't seem necessary to go into finer points. More precisely, if the second derivative exists (as it does in this case wherever the function is defined), it is a necessary condition for a point to be an inflection point that the second derivative vanishes. Thus this function has no inflection points. – joriki Feb 22 at 7:39
    
I think you should edit this correction into your answer. Comments are "second-class citizens" on Stack Exchange sites, and not everybody reads them. – Silverfish Feb 22 at 16:12
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@Silverfish: OK, I did. You're right -- though your comment is somewhat paradoxical, since you apparently read the comment :-) – joriki Feb 22 at 16:19

An inflection point is where the sign of curvature changes, from concave up to down or vice versa, hence the necessity of the vanishing second derivative.

But $f(x) = \frac{x^2}{2} - \ln x$ is concave up (convex) before and after the stationary point $x = 1$. This can be seen from the graph, or by noting that for all $x \in \mathbb{R}$ $$f''(x) = 1 + \frac{1}{x^2} > 0$$ So there can't be any inflection point.

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Counterexample: $x\mapsto x + x^3$. At $x = 0$ has a point of inflection with derivative $\ne 0$.

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Why is the derivative at $x=0$ for $f(x)=x^3 \ne0$? – GoodDeeds Feb 22 at 7:38
    
@GoodDeeds, corrected. – Martín-Blas Pérez Pinilla Feb 22 at 7:41

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