Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When f(x) = (x-2)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 A-C to sketch the graph.

- anonymous

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- anonymous

For the first derivative I got: f'(x) = e^x (x-1)

- DanJS

yes

- anonymous

D

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- anonymous

So 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

When the first derivative in fraction form, I got: e^x - (x-2)* (e^x) all over (e^-x)^2

- anonymous

so when f'(x) is undefined, does x= 0? Im not sure...

- DanJS

sorry my mouse died

- DanJS

yes x=1 has a horizontal tangent point, test either side of it to see what the function is doing

- anonymous

f'(x)=0 gives x=1

- DanJS

is f '(x) positive(increasing) or negative(decreasing) function on etiher side of x=1, test any point

- DanJS

Just for practice, here is the actual graph of f, to compare your answers to..

##### 1 Attachment

- DanJS

since x=1 is the only critical point here, you have two intervals to test for inc/dec
(-inf, 1) and (1, +inf)

- anonymous

Really? what about when f'(x) = undefined, does that give us any critical values?

- DanJS

where does that happen?

- anonymous

so 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

f ' (x) = (x-1)e^x
what did you do next

- anonymous

So 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 - (x-2)* (e^x) all over (e^-x)^2

- DanJS

Oh, do you mean the original function f(x)

- anonymous

Yea from the original function, I tried to find the first derivative in a fraction form..and well I got e^x - (x-2)* (e^x) all over (e^-x)^2

- DanJS

f(x) = (x-2)e^x = xe^x - 2e^x
f(x) = xe^x - 2e^x

- DanJS

i dont understand what you did

- DanJS

using the product rule here , no quotient

- anonymous

Okay, so I basically did: f(x)= (x-2)/ 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

If I took e^x down to become e^-x, wouldnt I be able to use the quotient rule? That seems logical to me

- DanJS

yes, that gets messy with ln functions

- DanJS

but it will come out to x=1 the same

- DanJS

when you set it to zero

- anonymous

I didnt knowthat when (e^-x)^2 =0 it solves out to be x=1

- DanJS

e^(-2x) never is zero

- anonymous

So we get out nothing when f'(x)= undefined? Either way I got inc from: (1,infinity) and decreasing from: (negative infinity, 1).

- DanJS

there is no undefined value for the derivative

- anonymous

Okay good. And for local max, i got none. But for local min I got x= 1

- anonymous

Then 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

And 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

|dw:1437606668108:dw|

- DanJS

yes so far so good

- DanJS

so far you have from first derivative...
|dw:1437606825963:dw|

- DanJS

you can also calculate the x and y intercepts from original function

- DanJS

and test the limits as x goes to + and - infinity for horizontal asmytotes

- anonymous

So I got for CU: (0,1) U (1,infinity) and CD: (negative infinity, 0)

- DanJS

looks righ

- DanJS

t

- anonymous

And for the inflection points, do I plug in zero into the second derivative equation or original equation>

- DanJS

original, you want to add it to the graph of F(X)

- anonymous

right

- anonymous

So I got for inflection points: (0,-2)

- DanJS

|dw:1437607095039:dw|

- DanJS

right, so that is what your graph looks kinda like with the info so far

- DanJS

maybe calculate an X intercept when f(x)=0

- DanJS

and test for any limits when x goes off to infinities

- anonymous

I put the original equation into my graphing calc, and it gave me something diff

- anonymous

|dw:1437607248105:dw|

- DanJS

yes, what is the limit as x goes off to neg infinity for the original function...

- DanJS

e^(x) becomes very very small when x is larger negative numbers

- anonymous

close to zero?

- DanJS

yes
\[\lim_{x \rightarrow - \infty}=0\]

- DanJS

so the function approaches zero as x goes off to negative infinity .. like you drew

- DanJS

plug 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 (x-2) to make it negative

- anonymous

Yea, totes see it! thanks

- DanJS

that'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

Nah, dont need that. Do you mind helping me go through another similar problem like this?

- DanJS

ill try. hehe

- DanJS

open new thread and close it , ill move over

- anonymous

kk

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