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?

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

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

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

okay i'll do that. Thanks bunches :D

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

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

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

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

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

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

yup you are right

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

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

Yeah, but something is weird about this

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

I think it's increasing from -1 to 3

Post the choices

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

Do you read what I posted above?

yea and i think your right its D

:o)

i think you made a mistake on the derivative

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

thank you your wisdom is much appreciated! :)

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

@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

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

What kind of graphing calc are you using?

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

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

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

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