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OpenStudy (anonymous):

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?

hero (hero):

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

OpenStudy (anonymous):

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

hero (hero):

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

OpenStudy (anonymous):

okay i'll do that. Thanks bunches :D

hero (hero):

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

hero (hero):

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

hero (hero):

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

OpenStudy (anonymous):

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

hero (hero):

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

hero (hero):

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

OpenStudy (anonymous):

yup you are right

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

hero (hero):

Yeah, but something is weird about this

OpenStudy (anonymous):

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

hero (hero):

I think it's increasing from -1 to 3

hero (hero):

Post the choices

OpenStudy (anonymous):

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

hero (hero):

Do you read what I posted above?

OpenStudy (anonymous):

yea and i think your right its D

hero (hero):

:o)

OpenStudy (anonymous):

i think you made a mistake on the derivative

hero (hero):

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

OpenStudy (anonymous):

thank you your wisdom is much appreciated! :)

hero (hero):

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

OpenStudy (anonymous):

@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

hero (hero):

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

hero (hero):

What kind of graphing calc are you using?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

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