# Taylor’s derivative tests for extrema and inflexion points

Hi everyone. 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 Note by J D
3 years, 3 months ago

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