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anonymous
 one year ago
Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When f(x) = (x2)e^x
A) Find the intervals of increase of decrease
B) Find the local max and mins
C) Find the intervals of concavity and inflection points
D) Use the information from parts AC to sketch the graph.
anonymous
 one year ago
Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When f(x) = (x2)e^x A) Find the intervals of increase of decrease B) Find the local max and mins C) Find the intervals of concavity and inflection points D) Use the information from parts AC to sketch the graph.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For the first derivative I got: f'(x) = e^x (x1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So for the critical values, for when f'(x) = 0 I got x=1. But for the other critical value when f'(X)= undefined... I tried to put the first derivative in a fraction form but Im not sure Its right.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0When the first derivative in fraction form, I got: e^x  (x2)* (e^x) all over (e^x)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so when f'(x) is undefined, does x= 0? Im not sure...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes x=1 has a horizontal tangent point, test either side of it to see what the function is doing

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0is f '(x) positive(increasing) or negative(decreasing) function on etiher side of x=1, test any point

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Just for practice, here is the actual graph of f, to compare your answers to..

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0since x=1 is the only critical point here, you have two intervals to test for inc/dec (inf, 1) and (1, +inf)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Really? what about when f'(x) = undefined, does that give us any critical values?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so you would have to do the first derivative in a fraction form and then whatever is at the bottom of that fraction, you set that to zero and solve...and whatever you get from that is another critical value. But i'm not even sure that I put the first derivative in its right fraction form

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0f ' (x) = (x1)e^x what did you do next

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I actually got what you got, but then I did it all over again trying to put it into fraction form and this is what I got: e^x  (x2)* (e^x) all over (e^x)^2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0Oh, do you mean the original function f(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea from the original function, I tried to find the first derivative in a fraction form..and well I got e^x  (x2)* (e^x) all over (e^x)^2

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0f(x) = (x2)e^x = xe^x  2e^x f(x) = xe^x  2e^x

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0i dont understand what you did

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0using the product rule here , no quotient

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so I basically did: f(x)= (x2)/ e^x and then I did the quotient rule on this to find the first derivative in fraction form But I assumed to use the quotient rule because I took the e^x down

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I took e^x down to become e^x, wouldnt I be able to use the quotient rule? That seems logical to me

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, that gets messy with ln functions

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0but it will come out to x=1 the same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I didnt knowthat when (e^x)^2 =0 it solves out to be x=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we get out nothing when f'(x)= undefined? Either way I got inc from: (1,infinity) and decreasing from: (negative infinity, 1).

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0there is no undefined value for the derivative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay good. And for local max, i got none. But for local min I got x= 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then to find the intervals of concavity and inflection points I know I have to find the second derivative and I got f''(x) = e^x (x) which basically gives me x=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And if I am right with the critical value... do I do a number line that has 1 and 0 and test out points between them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437606668108:dw

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0so far you have from first derivative... dw:1437606825963:dw

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0you can also calculate the x and y intercepts from original function

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0and test the limits as x goes to + and  infinity for horizontal asmytotes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I got for CU: (0,1) U (1,infinity) and CD: (negative infinity, 0)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And for the inflection points, do I plug in zero into the second derivative equation or original equation>

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0original, you want to add it to the graph of F(X)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I got for inflection points: (0,2)

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0right, so that is what your graph looks kinda like with the info so far

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0maybe calculate an X intercept when f(x)=0

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0and test for any limits when x goes off to infinities

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I put the original equation into my graphing calc, and it gave me something diff

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1437607248105:dw

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes, what is the limit as x goes off to neg infinity for the original function...

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0e^(x) becomes very very small when x is larger negative numbers

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0yes \[\lim_{x \rightarrow  \infty}=0\]

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0so the function approaches zero as x goes off to negative infinity .. like you drew

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0plug in smaller and smaller values for x in the original function and you can see how e^(x) behaves for smaller and smaller values of x, and it is multiplied by (x2) to make it negative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea, totes see it! thanks

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0that's about it.. less you want to calculate the x intercept and add that point, and show that the limit at positive infinity is infinity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nah, dont need that. Do you mind helping me go through another similar problem like this?

DanJS
 one year ago
Best ResponseYou've already chosen the best response.0open new thread and close it , ill move over
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