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Consider this function:

$$f(x)= 5x^6 - 18x^5 + 15x^4 - 10$$

I am told to find the extreme values of this function. So at first, I take the first derivative and set it zero.

$$f'(x)=30x^5-90x^4+60x^3=0(say)$$

$$=x^5-3x^4+2x^3=0$$

If I evaluate for $x$, I find $x= 0, 1, 2$. Now let's take the second derivative of $f(x)$:

$$f''(x)= 150x^4-360x^3+180x^2$$

I am Ok with $f''(1)$ and $f''(2)$ because they give a negative and positive value respectively. Now, $f''(0) = 0$. So I don't know if I have a maxima or minima at $x=0$. I have found one way to solve this on internet that is named as first derivative test. Here I take two values of $x$, one $<0$, other $>0$. Let me take $0.5$ and $-1$.

$$f'(-1)= -180$$

$$f'(0.5)=45/16$$

So if $x<0$ then the function is decreasing. But when $0 < x <1$ (less than one because we have another maxima at $x=1$, but I am worrying only about that at $x=0$), it's increasing. Thus I can say we have a minima at $x=0$.

The main reason behind my asking is, after finding $f''(x)=0$, my solution book has shown that $f^{(iii)}(0)=0$ but $f^{(iv)}(0)= 360>0$. Thus we have a minima at $x=0$. But this is the INFLECTION POINT test, isn't it? But a point being an inflection point doesn't necessarily mean that it is a local maxima or local minima. So I would like to know if my work is right as I am not deeply familiar with the solution that I gave above and if my book's solution is also allowable or not?

And one more thing I am not sure about: Can a point be an inflection point and at the same time a local maxima or minima? If yes, an example would be very helpful.

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It has a local minimum at $0$, because $f^{(3)}(0)=0$ and $f^{(4)}(0)>0$.

The general rule here is: you keep derivating until you get the first $n$ such that $f^{(n)}(a)\neq0$. Then:

  • if $n$ is odd, $a$ is neither a local maximum nor a local minimum;
  • if $n$ even and $f^{(n)}(a)>0$, then $f$ has a local minimum at $a$;
  • if $n$ even and $f^{(n)}(a)<0$, then $f$ has a local maximum at $a$.
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    $\begingroup$ This follows from Taylor's Theorem. $\endgroup$ – Wauzl Jan 4 '18 at 10:14
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    $\begingroup$ @Wauzl I know that. $\endgroup$ – José Carlos Santos Jan 4 '18 at 10:15
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    $\begingroup$ I'm pretty sure you knew that, I was just stating this for OP, in case he wondered where this result came from. $\endgroup$ – Wauzl Jan 4 '18 at 10:19
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    $\begingroup$ @Wauzl That was a nice thought. $\endgroup$ – José Carlos Santos Jan 4 '18 at 10:28
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    $\begingroup$ @Sami You should make an effort to understand why it works. It's not that hard. It all boils down to the fact that, if $n$ is as the same $n$ as in my answer, then the limit $\lim_{x\to a}\frac{f(x)-f(a)}{(x-a)^n}$ exists and it has the same sign as $f^{(n)}(a)$. $\endgroup$ – José Carlos Santos Jan 4 '18 at 10:33
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Close to zero, discarding the smaller terms, the function is essentially

$$15x^4-10,$$ with derivatives $60x^3$ and $180x^2$.

Clearly, the second derivative remains positive, the first derivative is increasing and changes sign (from negative to positive) and the function itself has a minimum, as it is decreasing then increasing.

The plot shows you the true function, the approximation at $x=0$ and its derivatives (I let you guess which is which).

enter image description here

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  • $\begingroup$ well i guess, the blue one is the true function, green is the approximation at x=0, pink one is the first derivative and the black one is the second derivative. And thanks for the nice and clear answer. $\endgroup$ – Sami Jan 4 '18 at 10:40
  • $\begingroup$ @Sami: you are welcome and right. $\endgroup$ – Yves Daoust Jan 4 '18 at 10:45
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Consider $f(x)$ whose first n derivations are zero in $x_0$. An even n implies that point is inflection point and an odd n implies it to be maximum or minimum. then to prove which case is happening exactly you need to calculate $f^{n+1}(x)$. Positivity of it implies minimum and negativity implies maximum

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Geometrically speaking, an inflection point is a point where the curve crosses its tangent line. The equation of the tangent line at $x_0$ is $y = f(x_0) + (x - x_0)f'(x_0)$. This is also true if $f'(x_0) = 0$. If the first non-zero derivative after the first one has odd order, then one has an inflection point (assuming such a derivative exists).

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For a point to be inflection point you need to have f'' has different signs on two sides of that point. Merely showing that f''(x)=0 does not guarantee an inflection point.

For your final question, it's a duplicate of Is it possible to be both a relative max/min and an inflection point?, check that post for a constructed function with inflection point and local extremum at that point at the same time.

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