f(x)=(x^2-3)e^-x What are all values of x for which f is increasing? A. There are no such values of x. B. X3 C. -3

Question

f(x)=(x^2-3)e^-x 
What are all values of x for which f is increasing? 
A. There are no such values of x.
B. X<-1 and x>3
C. -3<x<1
D. -1<x<3
E. all values of x.
I did f '(x)= 2xe^(-x)-(x^2-3)e^-x 
Now I have to set that equal to zero right? I'm stuck there.

zepdrix · Answer

Ok looks good so far.

zepdrix · Answer

$$\Large\bf\sf 0\quad=\quad e^{-x}\left[2x-(x^2-3)\right]$$

zepdrix · Answer

Factor e^{-x} out of each term.

zepdrix · Answer

We're looking for critical points, right?
That's the locations where the function changes from increasing to decreasing.
That's why I set it equal to zero.

Hopefully you already understand that though :)

darkigloo · Answer

yup

zepdrix · Answer

The exponential function (without a vectical shift) has `no x-intercepts`.

$\Large\bf\sf e^{-x}\ne0$

zepdrix · Answer

So we only need to work with the other factor.

zepdrix · Answer

$$\Large\bf\sf 0\quad=\quad 2x-(x^2-3)$$

darkigloo · Answer

x=-1 x=3

darkigloo · Answer

then i plug in a # between (-inf,-1), (-1,3), and (3, inf.) to find if f inc/dec right?

zepdrix · Answer

AHHH my browswer tweaked out :( sorry wrong window

zepdrix · Answer

lemme fix that.

darkigloo · Answer

no problem :)

zepdrix · Answer

Mmm yes, good! test points between those values :D

zepdrix · Answer

I like to draw the number line myself ^^
But whatever works for you.

darkigloo · Answer

alright i got it. thank you. :)

zepdrix · Answer

Cool :3

anonymous · Answer

Refer to the Mathematica attachment.