Question

What are all values of x for which the function f defined by f(x)=(x^2-3)e^-x is increasing?

A step by step explanation would be nice. I don't know whether I graph it or work it algebraically?

Answer 1

I think the best approach here is to graph it and find the regions where the graph is increasing

Answer 2

do i take the derivative and then graph it or just graph it as it is?

Answer 3

Both. Graph it as is to see where it is literally increasing. Graph the derivative to see how it corresponds

Answer 4

okay i'll do that. Thanks bunches :D

Answer 5

If you graph it, you'll see that there is a zero when the graph goes as the graph goes from negative to positive

Answer 6

Take the derivative, set it equal to zero, then solve for x will give you the exact value where it is zero.

Answer 7

Then the interval where it is increasing should be obvious at that point

Answer 8

you are probably supposed to take the derivative, then check where it is postive

Answer 9

Yes, I instructed her to find the derivative and graph it

Answer 10

I'm sure you read what I wrote, so....I don't believe I was misleading

Answer 11

yup you are right

Answer 12

when you take the derivative factor out the
\[e^{-x}\] term and just worry about the quadratic

Answer 13

so would my derivative be (e^-x)(2x)+(x^2-3)(e^-x) and then i take out the e^-x so that it will only be 2x+x^2-3=0

Answer 14

Yeah, but something is weird about this

Answer 15

it gives me multiple choice answers would you like for me to post those

Answer 16

I think it's increasing from -1 to 3

Answer 17

Post the choices

Answer 18

A) there are no such values of x
B) x<-1 and x>3
C) its not c cause i got it wrong
D) -1<x<3

Answer 19

Do you read what I posted above?

Answer 20

yea and i think your right its D

Answer 21

:o)

Answer 22

i think you made a mistake on the derivative

Answer 23

If you had just graphed the original function, you could see where it was increasing, just like I said originally.

Answer 24

thank you your wisdom is much appreciated! :)

Answer 25

In this case, you didn't really need the derivative

Answer 26

@hero it is easy to see from the graph that it starts increasing at -1, but not so easy to see where it starts decreasing

Answer 27

I looked at the graph and figured out that it was increasing between -1 and 3 without any further math.

Answer 28

What kind of graphing calc are you using?

Answer 29

the mistake in the derivative is that that the derivative of
\[e^{-x}\] is
\[-e^{-x}\] so that part of your polynomial is off by a sign

Answer 30

yea i see it now! thanks....my calulator is a TI-84 Plus

Answer 31

derivative is
\[e^{-x} (-x^2+2 x+3)\]