The derivative of f is given f'(x) = e^x(-x^3+3x)-3 for [0,5]. What value of f(x) is an absolute minimum?

Question

jennychan12 · Answer

$$f'(x) = e^x(-x^2+3x)-3 $$ for
$$0 \le x \le 5$$

jennychan12 · Answer

i used a graphing calculator to find the x-coordinate at the minimum value. is there a way to do this without a graphing calculator?

anonymous · Answer

The most you could do without a graphing calculator is set $f'(x)=0$ and try solving for $x$, as you normally would with the first derivative test. But solving for $x$ by hand would be very, very difficult.

anonymous · Answer

http://www.wolframalpha.com/input/?i=Roots%5Be%5Ex+%28-x%5E2%2B3x%29-3%3D0%5D

jennychan12 · Answer

yeah, i tried solving for x. it's kinda frustrating. and annoying. and this question is part of the no calculator, so i dunno how i would do it. :(

jennychan12 · Answer

thanks anyways :)