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)=x3e3x2 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 x=0

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

f(x)=(32x2)x2e3x2=0 f’(x) = (3 - 2x^2)x^2e^{3-x^2} = 0 Then the critical points are:

x=0 x=0

x=+32x= +-\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 f’’(0) = 0 A theorem tells us this result is inconclusive

f(32)=326e32<0 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=32x= -\sqrt{ \frac{3}{2}}

The prove:

f(32)=326e32>0 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 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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