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I am having little confusion regarding this graph , as we see that just before close to zero and to the right of zero the function changes its sign from negative to positive so then how come $x=0$ is a point of inflection. According to me $x=0$ should be a point of local minimum here .

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    $\begingroup$ Look again at the definition of local minimum. $\endgroup$ – Robert Israel Jul 12 '15 at 18:54
  • $\begingroup$ What makes you think it should be a minimum? (and the notes should have the singular minimum and maximum rather than minima and maxima - which are plural forms) $\endgroup$ – Mark Bennet Jul 12 '15 at 18:56
  • $\begingroup$ $0$ is greater than the neighborhood point $(x,y)$ which is in the third quadrant, so it's definitely not the local minimum. $\endgroup$ – MathGod Jul 12 '15 at 19:00
  • $\begingroup$ And is that book class 12th NCERT? $\endgroup$ – MathGod Jul 12 '15 at 19:01
  • $\begingroup$ The function $$y = \begin{cases} x^2 & x<0\\ \sqrt x & x\ge 0 \end{cases}$$ has a point of inflexion and local minimum at $x=0$. $\endgroup$ – peterwhy Jul 12 '15 at 19:02
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The first derivative of the function must take different signed values at two sides of the local minimum point.

This is not satisfied in the example above. Here, $(0,0)$ is an inflection point because $f''$ changes its sign at $x=0$.

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  • $\begingroup$ Is it necessary for the derivative to change sign at two sides of a local minimum? I don't think so. The derivative may oscillate that fast and change sign but still the function to obtain a local minimum or maximum. The answer to the question is no and it follows directly from the definitions. $\endgroup$ – Tolaso Jul 12 '15 at 19:22
  • $\begingroup$ Could you please me an example? $\endgroup$ – frosh Jul 12 '15 at 19:24
  • $\begingroup$ Have a look at this classic example: $$f(x)= \left\{\begin{matrix} x^2\sin \frac{1}{x} & , & x \neq 0\\ 0&, &x =0 \end{matrix}\right.$$ Both the derivative and the function oscillate near zero but still $f$ has a local minimum at $x=0$. $\endgroup$ – Tolaso Jul 12 '15 at 19:34
  • $\begingroup$ I updated my example since the function was not differentiable at $x=0$. $\endgroup$ – Tolaso Jul 12 '15 at 19:35
  • $\begingroup$ Another made up example, $$g(x) = \begin{cases} x& x<0\\ x-42 & x\ge 0 \end{cases}$$ has a local minimum at $x=0$. The function is not even continuous at $x=0$, but $g(0)$ is smaller than neighbouring $g(x)$'s, and the first derivative "at the two sides of the local minimum point" has the same sign. $\endgroup$ – peterwhy Jul 12 '15 at 19:47
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Let's try to clear things up.


At the graph you are given the graph of the function $f(x)=x^3$. It is apparently differentiable on $\mathbb{R}$ as polynomial with a derivative of $f'(x)=3x^2 \geq 0$. Note that $f'(x)=0 \Leftrightarrow x=0$. One could speculate here that $x=0$ is an extrema value for $f$. No, this is not the case because the derivative does not change sign around $0$.

Hence $f$ is stricly increasing on $\mathbb{R}$. Now $f'$ is again differentiable as a polynomial. Differentiating once more we get that $f''(x)=6x$. Now, we note that $f''(x)=0 \Leftrightarrow 6x =0 \Leftrightarrow x =0$.

We also see that around zero the second derivative changes sign , meaning that $f$ has an inflection point at $x=0$ according to a known theorem

Regarding your question if a function has an extrema value at $x_0$ then at that value cannot have an inflection point and vice versa.

Proof

I leave the details up to you, but you may begin from $f''$ and study the monotony of $f$. You'll also use the process of elimination. Also you will use Fermat's theorem since $x_0$ is an internal point.

P.S 1: If for a function $f$ holds that $f'(x_0)=0$ then $x_0$ is not necessarily an extrema value of $f$. See $f(x)=x^3$.

P.S 2: Same holds if $f''(x_0)=0$ then it is not necessary an inflection point.

Hope this clears things up a bit.

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  • $\begingroup$ I may be mistaken, but I don't think your proof works for $f(x)=x^4$ (since f has a local minimum at 0, but $f^{\prime\prime}(0)=0$). $\endgroup$ – user84413 Jul 12 '15 at 20:12
  • $\begingroup$ @user84413 Yes, you are right. I used the converse which does not hold necessariy. I'll try to fix the proof. I sure have one written down in my notebooks. $\endgroup$ – Tolaso Jul 12 '15 at 20:17

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