## darkigloo one year ago If f(x)=(x^2)(e^x) then f is decreasing for all x such that A. x<-2 B. -2<x<0 C. x>-2 D. x<0 E. x>0

1. myininaya

Calculus or algebra?

2. darkigloo

calculus

3. myininaya

Ok cool stuff. So do you know how to find the derivative. The derivative is the function that tells if the original function is decreasing or increasing.

4. darkigloo

i think the derivative is e^x(x^2+2x)

5. myininaya

that is right So when is the derivative 0 (this is to find when the function is doing neither increasing or increasing) I'm asking this: solve e^x(x^2+2x)=0

6. myininaya

e^x is never 0

7. myininaya

So you are basically looking at solving x^2+2x=0 right now.

8. darkigloo

x=0 x=-2

9. myininaya

ok great. Now lets check the intervals around those numbers to see if we have f'>0 (which means f is increasing) or if we have f'<0 (which means f is decreasing) So what I'm asking you to do is look at the interval (-inf,-2) Take a number from this set and see if f'>0 or <0. Then we will also check the set of numbers (-2,0) and also check (0,inf)

10. myininaya

we have three intervals to check.

11. myininaya

when ever you plug in a number into e^x all you need to know is that the output will always be positive

12. myininaya

So again looking at just x^2+2x

13. myininaya

A number from the group (-inf,-2), we could choose -4. (-4)^2+2(-4)=a positive number>0 so on this interval it is increasing. Now you check (-2,0).

14. darkigloo

for (-2,0) its negative. for (0,-inf) its positive. So the answer is B. -2<x<0 right!? :D

15. myininaya

that is right

16. darkigloo

thanks. and for the e^x...that is always positive?

17. myininaya

any positive number raised to any power will always be positve

18. darkigloo

ok. thank you.

19. myininaya

|dw:1393014251807:dw|

20. myininaya

Like you can also look at the graph if you want it gets closer to 0 as we approach negative infinity but never touches or goes below

21. darkigloo

ohh right...i'll have to remember that graph. thanks :)

22. myininaya