Question

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.

Answer 1

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

Answer 2

yes

Answer 3

D

Answer 4

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.

Answer 5

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

Answer 6

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

Answer 7

sorry my mouse died

Answer 8

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

Answer 9

f'(x)=0 gives x=1

Answer 10

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

Answer 11

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

Answer 12

since x=1 is the only critical point here, you have two intervals to test for inc/dec

(-inf, 1) and (1, +inf)

Answer 13

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

Answer 14

where does that happen?

Answer 15

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

Answer 16

f ' (x) = (x-1)e^x

what did you do next

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

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

Answer 21

i dont understand what you did

Answer 22

using the product rule here , no quotient

Answer 23

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

Answer 24

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

Answer 25

yes, that gets messy with ln functions

Answer 26

but it will come out to x=1 the same

Answer 27

when you set it to zero

Answer 28

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

Answer 29

e^(-2x) never is zero

Answer 30

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

Answer 31

there is no undefined value for the derivative

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

yes so far so good

Answer 36

so far you have from first derivative...