anonymous
  • anonymous
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.
Calculus1
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
For the first derivative I got: f'(x) = e^x (x-1)
DanJS
  • DanJS
yes
anonymous
  • anonymous
D

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
  • anonymous
When the first derivative in fraction form, I got: e^x - (x-2)* (e^x) all over (e^-x)^2
anonymous
  • anonymous
so when f'(x) is undefined, does x= 0? Im not sure...
DanJS
  • DanJS
sorry my mouse died
DanJS
  • DanJS
yes x=1 has a horizontal tangent point, test either side of it to see what the function is doing
anonymous
  • anonymous
f'(x)=0 gives x=1
DanJS
  • DanJS
is f '(x) positive(increasing) or negative(decreasing) function on etiher side of x=1, test any point
DanJS
  • DanJS
Just for practice, here is the actual graph of f, to compare your answers to..
1 Attachment
DanJS
  • 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
  • anonymous
Really? what about when f'(x) = undefined, does that give us any critical values?
DanJS
  • DanJS
where does that happen?
anonymous
  • 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
  • DanJS
f ' (x) = (x-1)e^x what did you do next
anonymous
  • 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
  • DanJS
Oh, do you mean the original function f(x)
anonymous
  • 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
  • DanJS
f(x) = (x-2)e^x = xe^x - 2e^x f(x) = xe^x - 2e^x
DanJS
  • DanJS
i dont understand what you did
DanJS
  • DanJS
using the product rule here , no quotient
anonymous
  • 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
  • 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
  • DanJS
yes, that gets messy with ln functions
DanJS
  • DanJS
but it will come out to x=1 the same
DanJS
  • DanJS
when you set it to zero
anonymous
  • anonymous
I didnt knowthat when (e^-x)^2 =0 it solves out to be x=1
DanJS
  • DanJS
e^(-2x) never is zero
anonymous
  • 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
  • DanJS
there is no undefined value for the derivative
anonymous
  • anonymous
Okay good. And for local max, i got none. But for local min I got x= 1
anonymous
  • 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
  • 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
  • anonymous
|dw:1437606668108:dw|
DanJS
  • DanJS
yes so far so good
DanJS
  • DanJS
so far you have from first derivative... |dw:1437606825963:dw|
DanJS
  • DanJS
you can also calculate the x and y intercepts from original function
DanJS
  • DanJS
and test the limits as x goes to + and - infinity for horizontal asmytotes
anonymous
  • anonymous
So I got for CU: (0,1) U (1,infinity) and CD: (negative infinity, 0)
DanJS
  • DanJS
looks righ
DanJS
  • DanJS
t
anonymous
  • anonymous
And for the inflection points, do I plug in zero into the second derivative equation or original equation>
DanJS
  • DanJS
original, you want to add it to the graph of F(X)
anonymous
  • anonymous
right
anonymous
  • anonymous
So I got for inflection points: (0,-2)
DanJS
  • DanJS
|dw:1437607095039:dw|
DanJS
  • DanJS
right, so that is what your graph looks kinda like with the info so far
DanJS
  • DanJS
maybe calculate an X intercept when f(x)=0
DanJS
  • DanJS
and test for any limits when x goes off to infinities
anonymous
  • anonymous
I put the original equation into my graphing calc, and it gave me something diff
anonymous
  • anonymous
|dw:1437607248105:dw|
DanJS
  • DanJS
yes, what is the limit as x goes off to neg infinity for the original function...
DanJS
  • DanJS
e^(x) becomes very very small when x is larger negative numbers
anonymous
  • anonymous
close to zero?
DanJS
  • DanJS
yes \[\lim_{x \rightarrow - \infty}=0\]
DanJS
  • DanJS
so the function approaches zero as x goes off to negative infinity .. like you drew
DanJS
  • 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
  • anonymous
Yea, totes see it! thanks
DanJS
  • 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
  • anonymous
Nah, dont need that. Do you mind helping me go through another similar problem like this?
DanJS
  • DanJS
ill try. hehe
DanJS
  • DanJS
open new thread and close it , ill move over
anonymous
  • anonymous
kk

Looking for something else?

Not the answer you are looking for? Search for more explanations.