[Math] Taylor’s derivative test for extrema and inflexion points

calculusmaxima-minimataylor expansion

I’m struggling to completely solve the following exercise:

• Study the function $f(x) = x^3e^{3-x^2}$ finding its solution/s, critical point/s and inflexion point/s.

Firstly, the unique solution is $x=0$

Secondly, to find the critical points we have to derivate the function and make it equal to 0. Therefore:

$f’(x) = (3 – 2x^2)x^2e^{3-x^2} = 0$
Then the critical points are:

$ x=0 $

$ x= +-\sqrt{ \frac{3}{2}} $

As we know, we have to do a second derivative and calculate it in function of the critical points’ value, in order to find the inflexion, maximum and minimum point/s.
Therefore:

$ f’’(0) = 0 $
A theorem tells us this result is inconclusive

$ f’’( \sqrt{ \frac{3}{2}}) = -\sqrt{ \frac{3}{2}} 6e^{ \frac{3}{2} } < 0 $

Then we have a maximum

As the function is odd we can directly claim than we have a minimum at the critical point :

$ x= -\sqrt{ \frac{3}{2}} $

The prove:

$ f’’( -\sqrt{ \frac{3}{2}}) = \sqrt{ \frac{3}{2}} 6e^{ \frac{3}{2} } > 0 $

Therefore we have a minimum

Once I arrive here, my struggle starts. I know that there is a method using Taylor to analyse the case:

$ f’’(0) = 0 $

It allows us to determine if there’s such an inflexion point and, what’s more, find out how can be drawn.

I have looked for this method in many books: Calculus of Marsden and Weinstein, Stewart…

But I find it nowhere.

Could someone explain me how to use the method and finish this problem?

if it is not possible to explain here in a detailed way, could you provide me a reference where I could understand it deeply?

Thank you very much

Best Answer

Note that to determine the nature of the critical points we can also study the sign of $f'(x)$ in the intervals that is

  • $f'(x)<0$ for $x<-\sqrt{ \frac{3}{2}}$
  • $f'(x)\ge0$ for $-\sqrt{ \frac{3}{2}}<x<\sqrt{ \frac{3}{2}}$
  • $f'(x)<0$ for $x>\sqrt{ \frac{3}{2}}$

and thus $x=-\sqrt{ \frac{3}{2}}$ is apoint of minimum and $x=\sqrt{ \frac{3}{2}}$ is a point of maximum.

To determine the nature of the point $x=0$ we can consider the sign of $f''(x)$ in a neighborhood of $x=0$ or as an alternative we need to derive further in order to find the first derivative of order $k$ for which $f^k(0)\neq 0$. If $k$ is odd we have proved that at $x=0$ we have an inflection point.

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