Absolute Maximum of f(x)

I have the function f(x) = (x^3) / (e^x), e as in euler's number.
I'm not sure how to find the absolute maximum; i derived it and found critical values, but the main problem is that i ended up with 0,0 even though the graph clearly goes above that point. Any insight or help is appreciated.

I am getting that 27/e^3 is the global maximum. Maybe you did something wrong?

With f(x) = x^3/e^x = x^3*e^(-x), we have that:
f'(x) = 3x^2*e^(-x) - x^3*e^(-x)
= e^(-x)(3x^2 - x^3), by factoring out e^(-x)
= x^2*e^(-x)(3 - x).

Setting f'(x) = 0 gives:
x^2*e^(-x)(3 - x) = 0 ==> x = 0 and x = 3, by the zero-product property.
(Note that e^(-x) > 0 for all x.)

Using the Second Derivative Test, you can show that x = 0 is actually an inflection point (since f''(0) = 0). x = 3 produces the desired maximum of f(3) = 27/e^3.

I hope this helps!

0 would be a saddle point, a point where the graph flattens out, but doesn't actually change direction. Basically, the graph stops rising and then starts rising again (or stops dropping and then starts dropping again). I believe the derivative is 3*x^2*e^(-x) - x^3*e^(-x). Setting this equal to 0 gives two solutions: 3 and 0. As I explained, 0 is a saddle point. Looking at 3 however, we can see that values for x which are smaller that 3 as well as values for x which are larger than 3 give values of f(x) which are less that what you get when you plug 3 in to f(x) (which, by the way is 1.344...)

Hello,

f(x) = x³ / e^x

f'(x) = (3x².e^x - x³.e^x) / e^(2x) = (3x² - x³) / e^x

The derivative is obviously nil when x=0 and x=3.

f(0) = 0
f(3) = 27/e³ ≈ 1.344

If x<0, f'(x)>0 and f is increasing.
If 00 and f is increasing.
If 3
Thus f(3)=27/e³ is the absolute maximum.

Regards,
Dragon.Jade ;-)