Is there any other way to solve it?

[vilhelm]

Homework Statement

f(x)=x^k * e^-x
where k>1 and natural.
analyse how the coordinates of max, min and * depends on k.
Proove that your conclusion holds for any natural k, larger than 1.

*I'm not fluent in english, so I don't know the word, but it is what is called in (0,0) on for example the curve x^3 (i.e. the derivative is positive, zero, then positive or neg, zero, neg)

Homework Equations

The Attempt at a Solution

Homework Statement

Homework Equations

The Attempt at a Solution

 

[vilhelm]

I've solved almost everything, except:
why has f(x) an * if k is odd and a min if k is even?

 

[HallsofIvy]

I suspect your "*" represents "inflection point" where the second derivative changes sign- which, of course , can only occur where the second derivative is 0.

The derivative of f is
$$f'= kx^{k-1}e^{-x}- x^ke^{-x}= (k- x)x^{k-1}e^{-x}$$
and the second derivative is
$$f''= k(k-1)x^{k-2}e^{-x}- 2kx^{k-1}e^{-x}+ x^ke^{-x}= (k(k-1)- 2kx+ x^2)x^{k-2}e^{-k}$$

Now that exponential, $e^{-k}$ is never 0 so the second derivative is 0 only where $(x^2- 2kx+ k(k-1))x^{k-2}= ((x- k)^2- k)x^{k-2}= 0$. Obviously that second derivative is 0 at x= 0 but to determine whether or not it changes sign at x= 0, we have to look at $x^{k-2}$. If k is even, k-2 is also and $x^{k-2}$ is positive on both sides of x= 0: the second derivative does not change sign so 0 is NOT a inflection point. If k is odd, k- 2 is also and $x^{k-2}$ is positive for x> 0 and negative for x< 0: the second derivative changes sign so 0 is an inflection point.

By the way, the second derivative will also be 0 at $x= k\pm\sqrt{k}$ so there can be other inflection points.

 

[vilhelm]

I suspect your "*" represents "inflection point" where the second derivative changes sign- which, of course , can only occur where the second derivative is 0.

The derivative of f is
$$f'= kx^{k-1}e^{-x}- x^ke^{-x}= (k- x)x^{k-1}e^{-x}$$
and the second derivative is
$$f''= k(k-1)x^{k-2}e^{-x}- 2kx^{k-1}e^{-x}+ x^ke^{-x}= (k(k-1)- 2kx+ x^2)x^{k-2}e^{-k}$$

Now that exponential, $e^{-k}$ is never 0 so the second derivative is 0 only where $(x^2- 2kx+ k(k-1))x^{k-2}= ((x- k)^2- k)x^{k-2}= 0$. Obviously that second derivative is 0 at x= 0 but to determine whether or not it changes sign at x= 0, we have to look at $x^{k-2}$. If k is even, k-2 is also and $x^{k-2}$ is positive on both sides of x= 0: the second derivative does not change sign so 0 is NOT a inflection point. If k is odd, k- 2 is also and $x^{k-2}$ is positive for x> 0 and negative for x< 0: the second derivative changes sign so 0 is an inflection point.

By the way, the second derivative will also be 0 at $x= k\pm\sqrt{k}$ so there can be other inflection points.

Thanks! However, I solved it and it looks like we have similar solutions (great).